:: Sets and Functions of Trees and Joining Operations of Trees
:: by Grzegorz Bancerek
::
:: Received November 27, 1992
:: Copyright (c) 1992-2018 Association of Mizar Users
:: (Stowarzyszenie Uzytkownikow Mizara, Bialystok, Poland).
:: This code can be distributed under the GNU General Public Licence
:: version 3.0 or later, or the Creative Commons Attribution-ShareAlike
:: License version 3.0 or later, subject to the binding interpretation
:: detailed in file COPYING.interpretation.
:: See COPYING.GPL and COPYING.CC-BY-SA for the full text of these
:: licenses, or see http://www.gnu.org/licenses/gpl.html and
:: http://creativecommons.org/licenses/by-sa/3.0/.
environ
vocabularies NUMBERS, SUBSET_1, FINSEQ_1, FUNCT_1, XXREAL_0, ORDINAL4,
RELAT_1, TREES_1, XBOOLE_0, FINSET_1, TARSKI, TREES_2, FUNCT_2, CARD_1,
FINSEQ_2, FUNCOP_1, FUNCT_6, ZFMISC_1, PARTFUN1, MCART_1, NAT_1, ARYTM_3,
TREES_A, ARYTM_1, TREES_3;
notations TARSKI, XBOOLE_0, ZFMISC_1, SUBSET_1, ORDINAL1, NUMBERS, XCMPLX_0,
NAT_1, RELAT_1, FUNCT_1, FINSEQ_1, FINSET_1, DOMAIN_1, FUNCOP_1,
RELSET_1, PARTFUN1, BINOP_1, FUNCT_2, FUNCT_3, FINSEQ_2, TREES_1,
TREES_2, FUNCT_6, XXREAL_0;
constructors PARTFUN1, BINOP_1, DOMAIN_1, FUNCT_3, SQUARE_1, NAT_1, FINSEQ_2,
FUNCT_6, TREES_2, RELSET_1;
registrations XBOOLE_0, SUBSET_1, RELAT_1, FUNCT_1, ORDINAL1, FUNCT_2,
FINSET_1, XXREAL_0, XREAL_0, NAT_1, FINSEQ_1, TREES_1, TREES_2, CARD_1,
RELSET_1, FINSEQ_2;
requirements NUMERALS, REAL, BOOLE, SUBSET, ARITHM;
definitions TARSKI, XBOOLE_0, FUNCT_1, TREES_1, TREES_2, FUNCOP_1;
equalities TREES_1, TREES_2, BINOP_1;
expansions TARSKI, XBOOLE_0, FUNCT_1, TREES_1, TREES_2;
theorems TARSKI, ZFMISC_1, FINSEQ_1, MCART_1, NAT_1, FUNCT_1, FUNCT_2,
FUNCT_3, FUNCOP_1, FUNCT_6, FINSEQ_2, FINSEQ_3, ENUMSET1, TREES_1,
TREES_2, RELAT_1, RELSET_1, XBOOLE_0, XBOOLE_1, XREAL_1, XXREAL_0,
ORDINAL1, CARD_1;
schemes NAT_1, CARD_2, XBOOLE_0, FINSEQ_1;
begin :: Finite sets
reserve x,y,z for object, X,Y for set,
i,k,n for Nat,
p,q,r,s for FinSequence,
w for FinSequence of NAT,
f for Function;
Lm1: 1 <= n & n <= len p implies (p^q).n = p.n
proof
assume that
A1: 1 <= n and
A2: n <= len p;
n in dom p by A1,A2,FINSEQ_3:25;
hence thesis by FINSEQ_1:def 7;
end;
begin :: Sets of trees
definition
func Trees -> set means
: Def1:
x in it iff x is Tree;
existence
proof
set X = {T where T is Subset of NAT*: T is Tree};
take X;
let x;
thus x in X implies x is Tree
proof
assume x in X;
then ex T being Subset of NAT* st x = T & T is Tree;
hence thesis;
end;
assume x is Tree;
then reconsider T = x as Tree;
T is Subset of NAT* by TREES_1:def 3;
hence thesis;
end;
uniqueness
proof
defpred X[object] means $1 is Tree;
let T1,T2 be set such that
A1: x in T1 iff X[x] and
A2: x in T2 iff X[x];
thus thesis from XBOOLE_0:sch 2(A1,A2);
end;
end;
registration
cluster Trees -> non empty;
coherence
proof
set T = the Tree;
T in Trees by Def1;
hence thesis;
end;
end;
definition
func FinTrees -> Subset of Trees means
: Def2:
x in it iff x is finite Tree;
existence
proof
set X = {T where T is Element of Trees: T is finite Tree};
X c= Trees
proof
let x be object;
assume x in X;
then ex T being Element of Trees st x = T & T is finite Tree;
hence thesis;
end;
then reconsider X as Subset of Trees;
take X;
let x;
thus x in X implies x is finite Tree
proof
assume x in X;
then ex T being Element of Trees st x = T & T is finite Tree;
hence thesis;
end;
assume x is finite Tree;
then reconsider T = x as finite Tree;
T in Trees by Def1;
hence thesis;
end;
uniqueness
proof
defpred X[object] means $1 is finite Tree;
let T1,T2 be Subset of Trees such that
A1: x in T1 iff X[x] and
A2: x in T2 iff X[x];
thus thesis from XBOOLE_0:sch 2(A1,A2);
end;
end;
registration
cluster FinTrees -> non empty;
coherence
proof
set T = the finite Tree;
T in FinTrees by Def2;
hence thesis;
end;
end;
definition
let IT be set;
attr IT is constituted-Trees means
: Def3:
for x st x in IT holds x is Tree;
attr IT is constituted-FinTrees means
:
Def4: for x st x in IT holds x is finite Tree;
attr IT is constituted-DTrees means
:
Def5: for x st x in IT holds x is DecoratedTree;
end;
theorem
X is constituted-Trees iff X c= Trees
proof
thus X is constituted-Trees implies X c= Trees
proof
assume
A1: for x st x in X holds x is Tree;
let x be object;
assume x in X;
then x is Tree by A1;
hence thesis by Def1;
end;
assume
A2: X c= Trees;
let x;
assume x in X;
hence thesis by A2,Def1;
end;
theorem
X is constituted-FinTrees iff X c= FinTrees
proof
thus X is constituted-FinTrees implies X c= FinTrees
proof
assume
A1: for x st x in X holds x is finite Tree;
let x be object;
assume x in X;
then x is finite Tree by A1;
hence thesis by Def2;
end;
assume
A2: X c= FinTrees;
let x;
assume x in X;
hence thesis by A2,Def2;
end;
theorem Th3:
X is constituted-Trees & Y is constituted-Trees iff X \/
Y is constituted-Trees
proof
thus X is constituted-Trees & Y is constituted-Trees implies
X \/ Y is constituted-Trees
proof
assume that
A1: for x st x in X holds x is Tree and
A2: for x st x in Y holds x is Tree;
let x;
assume x in X \/ Y;
then x in X or x in Y by XBOOLE_0:def 3;
hence thesis by A1,A2;
end;
assume
A3: for x st x in X \/ Y holds x is Tree;
thus X is constituted-Trees
proof
let x;
assume x in X;
then x in X \/ Y by XBOOLE_0:def 3;
hence thesis by A3;
end;
let x;
assume x in Y;
then x in X \/ Y by XBOOLE_0:def 3;
hence thesis by A3;
end;
theorem
X is constituted-Trees & Y is constituted-Trees implies
X \+\ Y is constituted-Trees
proof
assume that
A1: for x st x in X holds x is Tree and
A2: for x st x in Y holds x is Tree;
let x;
assume x in X \+\ Y;
then not x in X iff x in Y by XBOOLE_0:1;
hence thesis by A1,A2;
end;
theorem
X is constituted-Trees implies X /\ Y is constituted-Trees &
Y /\ X is constituted-Trees & X \ Y is constituted-Trees
proof
assume
A1: for x st x in X holds x is Tree;
thus X /\ Y is constituted-Trees
proof
let x;
assume x in X /\ Y;
then x in X by XBOOLE_0:def 4;
hence thesis by A1;
end;
hence Y /\ X is constituted-Trees;
let x;
assume x in X \ Y;
hence thesis by A1;
end;
theorem Th6:
X is constituted-FinTrees & Y is constituted-FinTrees iff
X \/ Y is constituted-FinTrees
proof
thus X is constituted-FinTrees & Y is constituted-FinTrees implies
X \/ Y is constituted-FinTrees
proof
assume that
A1: for x st x in X holds x is finite Tree and
A2: for x st x in Y holds x is finite Tree;
let x;
assume x in X \/ Y;
then x in X or x in Y by XBOOLE_0:def 3;
hence thesis by A1,A2;
end;
assume
A3: for x st x in X \/ Y holds x is finite Tree;
thus X is constituted-FinTrees
proof
let x;
assume x in X;
then x in X \/ Y by XBOOLE_0:def 3;
hence thesis by A3;
end;
let x;
assume x in Y;
then x in X \/ Y by XBOOLE_0:def 3;
hence thesis by A3;
end;
theorem
X is constituted-FinTrees & Y is constituted-FinTrees implies
X \+\ Y is constituted-FinTrees
proof
assume that
A1: for x st x in X holds x is finite Tree and
A2: for x st x in Y holds x is finite Tree;
let x;
assume x in X \+\ Y;
then not x in X iff x in Y by XBOOLE_0:1;
hence thesis by A1,A2;
end;
theorem
X is constituted-FinTrees implies X /\ Y is constituted-FinTrees &
Y /\ X is constituted-FinTrees & X \ Y is constituted-FinTrees
proof
assume
A1: for x st x in X holds x is finite Tree;
thus X /\ Y is constituted-FinTrees
proof
let x;
assume x in X /\ Y;
then x in X by XBOOLE_0:def 4;
hence thesis by A1;
end;
hence Y /\ X is constituted-FinTrees;
let x;
assume x in X \ Y;
hence thesis by A1;
end;
theorem Th9:
X is constituted-DTrees & Y is constituted-DTrees iff
X \/ Y is constituted-DTrees
proof
thus X is constituted-DTrees & Y is constituted-DTrees implies
X \/ Y is constituted-DTrees
proof
assume that
A1: for x st x in X holds x is DecoratedTree and
A2: for x st x in Y holds x is DecoratedTree;
let x;
assume x in X \/ Y;
then x in X or x in Y by XBOOLE_0:def 3;
hence thesis by A1,A2;
end;
assume
A3: for x st x in X \/ Y holds x is DecoratedTree;
thus X is constituted-DTrees
proof
let x;
assume x in X;
then x in X \/ Y by XBOOLE_0:def 3;
hence thesis by A3;
end;
let x;
assume x in Y;
then x in X \/ Y by XBOOLE_0:def 3;
hence thesis by A3;
end;
theorem
X is constituted-DTrees & Y is constituted-DTrees implies
X \+\ Y is constituted-DTrees
proof
assume that
A1: for x st x in X holds x is DecoratedTree and
A2: for x st x in Y holds x is DecoratedTree;
let x;
assume x in X \+\ Y;
then not x in X iff x in Y by XBOOLE_0:1;
hence thesis by A1,A2;
end;
theorem
X is constituted-DTrees implies X /\ Y is constituted-DTrees &
Y /\ X is constituted-DTrees & X \ Y is constituted-DTrees
proof
assume
A1: for x st x in X holds x is DecoratedTree;
thus X /\ Y is constituted-DTrees
proof
let x;
assume x in X /\ Y;
then x in X by XBOOLE_0:def 4;
hence thesis by A1;
end;
hence Y /\ X is constituted-DTrees;
let x;
assume x in X \ Y;
hence thesis by A1;
end;
registration
cluster empty ->
constituted-Trees constituted-FinTrees constituted-DTrees for set;
coherence;
end;
theorem Th12:
{x} is constituted-Trees iff x is Tree
proof
thus {x} is constituted-Trees implies x is Tree
proof
assume
A1: for y st y in {x} holds y is Tree;
x in {x} by TARSKI:def 1;
hence thesis by A1;
end;
assume
A2: x is Tree;
let y;
thus thesis by A2,TARSKI:def 1;
end;
theorem Th13:
for x being object holds
{x} is constituted-FinTrees iff x is finite Tree
proof let x be object;
thus {x} is constituted-FinTrees implies x is finite Tree
proof
assume
A1: for y st y in {x} holds y is finite Tree;
x in {x} by TARSKI:def 1;
hence thesis by A1;
end;
assume
A2: x is finite Tree;
let y;
thus thesis by A2,TARSKI:def 1;
end;
theorem Th14:
{x} is constituted-DTrees iff x is DecoratedTree
proof
thus {x} is constituted-DTrees implies x is DecoratedTree
proof
assume
A1: for y st y in {x} holds y is DecoratedTree;
x in {x} by TARSKI:def 1;
hence thesis by A1;
end;
assume
A2: x is DecoratedTree;
let y;
thus thesis by A2,TARSKI:def 1;
end;
theorem Th15:
{x,y} is constituted-Trees iff x is Tree & y is Tree
proof
thus {x,y} is constituted-Trees implies x is Tree & y is Tree
proof
assume
A1: for z st z in {x,y} holds z is Tree;
A2: x in {x,y} by TARSKI:def 2;
y in {x,y} by TARSKI:def 2;
hence thesis by A1,A2;
end;
assume that
A3: x is Tree and
A4: y is Tree;
let z;
thus thesis by A3,A4,TARSKI:def 2;
end;
theorem Th16:
{x,y} is constituted-FinTrees iff x is finite Tree & y is finite Tree
proof
thus {x,y} is constituted-FinTrees implies x is finite Tree & y is
finite Tree
proof
assume
A1: for z st z in {x,y} holds z is finite Tree;
A2: x in {x,y} by TARSKI:def 2;
y in {x,y} by TARSKI:def 2;
hence thesis by A1,A2;
end;
assume that
A3: x is finite Tree and
A4: y is finite Tree;
let z;
thus thesis by A3,A4,TARSKI:def 2;
end;
theorem Th17:
{x,y} is constituted-DTrees iff x is DecoratedTree & y is DecoratedTree
proof
thus {x,y} is constituted-DTrees implies
x is DecoratedTree & y is DecoratedTree
proof
assume
A1: for z st z in {x,y} holds z is DecoratedTree;
A2: x in {x,y} by TARSKI:def 2;
y in {x,y} by TARSKI:def 2;
hence thesis by A1,A2;
end;
assume that
A3: x is DecoratedTree and
A4: y is DecoratedTree;
let z;
thus thesis by A3,A4,TARSKI:def 2;
end;
theorem
X is constituted-Trees & Y c= X implies Y is constituted-Trees;
theorem
X is constituted-FinTrees & Y c= X implies Y is constituted-FinTrees;
theorem
X is constituted-DTrees & Y c= X implies Y is constituted-DTrees;
registration
cluster finite constituted-Trees constituted-FinTrees non empty for set;
existence
proof set T = the finite Tree;
take {T};
thus thesis by Th12,Th13;
end;
cluster finite constituted-DTrees non empty for set;
existence
proof set T = the DecoratedTree;
take {T};
thus thesis by Th14;
end;
end;
registration
cluster constituted-FinTrees -> constituted-Trees for set;
coherence;
end;
registration
let X be constituted-Trees set;
cluster -> constituted-Trees for Subset of X;
coherence
by Def3;
end;
registration
let X be constituted-FinTrees set;
cluster -> constituted-FinTrees for Subset of X;
coherence
by Def4;
end;
registration
let X be constituted-DTrees set;
cluster -> constituted-DTrees for Subset of X;
coherence
by Def5;
end;
registration
let D be constituted-Trees non empty set;
cluster -> non empty Tree-like for Element of D;
coherence by Def3;
end;
registration
let D be constituted-FinTrees non empty set;
cluster -> finite for Element of D;
coherence by Def4;
end;
registration
cluster constituted-DTrees -> functional for set;
coherence;
end;
registration
let D be constituted-DTrees non empty set;
cluster -> DecoratedTree-like for Element of D;
coherence by Def5;
end;
registration
cluster Trees -> constituted-Trees;
coherence
by Def1;
end;
registration
cluster FinTrees -> constituted-FinTrees;
coherence
by Def2;
end;
registration
cluster constituted-FinTrees non empty for Subset of Trees;
existence
proof
take FinTrees;
thus thesis;
end;
end;
definition
let D be non empty set;
mode DTree-set of D -> non empty set means
:Def6: for x st x in it holds x is DecoratedTree of D;
existence
proof
take {(elementary_tree 0) --> the Element of D};
thus thesis by TARSKI:def 1;
end;
end;
registration
let D be non empty set;
cluster -> constituted-DTrees for DTree-set of D;
coherence
by Def6;
end;
registration
let D be non empty set;
cluster finite non empty for DTree-set of D;
existence
proof
set X = {the DecoratedTree of D};
X is constituted-DTrees
by TARSKI:def 1;
then reconsider X as non empty constituted-DTrees set;
X is DTree-set of D
proof
let x;
thus thesis by TARSKI:def 1;
end;
then reconsider X as DTree-set of D;
take X;
thus thesis;
end;
end;
registration
let D be non empty set, E be non empty DTree-set of D;
cluster -> D-valued for Element of E;
coherence by Def6;
end;
definition
let T be Tree, D be non empty set;
redefine func Funcs(T,D) -> non empty DTree-set of D;
coherence
proof
set F = Funcs(T,D);
F is constituted-DTrees
proof
let x;
assume x in F;
then ex f being Function st ( x = f)&( dom f = T)&( rng f c= D)
by FUNCT_2:def 2;
hence thesis by TREES_2:def 8;
end;
then reconsider F as non empty constituted-DTrees set;
F is DTree-set of D
proof
let x;
assume x in F;
then ex f being Function st ( x = f)&( dom f = T)&( rng f c= D)
by FUNCT_2:def 2;
hence thesis by RELAT_1:def 19,TREES_2:def 8;
end;
then reconsider F as DTree-set of D;
set d = the Function of T,D;
A1: dom d = T by FUNCT_2:def 1;
rng d c= D;
then d in F by A1,FUNCT_2:def 2;
hence thesis;
end;
end;
registration
let T be Tree, D be non empty set;
cluster -> DecoratedTree-like for Function of T,D;
coherence
by FUNCT_2:def 1;
end;
definition
let D be non empty set;
func Trees(D) -> DTree-set of D means
:Def7: for T being DecoratedTree of D holds T in it;
existence
proof
set A = the set of all Funcs(T,D) where T is Element of Trees;
set TT = union A;
set f = (the Element of Trees) --> the Element of D;
A1: f in Funcs(the Element of Trees,D) by FUNCT_2:8;
A2: Funcs(the Element of Trees,D) in A;
TT is constituted-DTrees
proof
let x;
assume x in TT;
then consider X such that
A3: x in X and
A4: X in the set of all Funcs(T,D) where T is Element of Trees
by TARSKI:def 4;
ex T being Element of Trees st ( X = Funcs(T,D)) by A4;
hence thesis by A3;
end;
then reconsider TT as non empty constituted-DTrees set
by A2,A1,TARSKI:def 4;
TT is DTree-set of D
proof
let x;
assume x in TT;
then consider X such that
A5: x in X and
A6: X in the set of all Funcs(T,D) where T is Element of Trees
by TARSKI:def 4;
ex T being Element of Trees st ( X = Funcs(T,D)) by A6;
hence thesis by A5;
end;
then reconsider TT as DTree-set of D;
take TT;
let T be DecoratedTree of D;
reconsider t = dom T as Element of Trees by Def1;
rng T c= D;
then
A7: T in Funcs(t,D) by FUNCT_2:def 2;
Funcs(t,D) in the set of all Funcs(W,D) where W is Element of Trees;
hence thesis by A7,TARSKI:def 4;
end;
uniqueness
proof
let T1,T2 be DTree-set of D such that
A8: for T being DecoratedTree of D holds T in T1 and
A9: for T being DecoratedTree of D holds T in T2;
thus T1 c= T2
by A9;
let x be object;
thus thesis by A8;
end;
end;
registration
let D be non empty set;
cluster Trees(D) -> non empty;
coherence;
end;
definition
let D be non empty set;
func FinTrees(D) -> DTree-set of D means
:Def8:
for T being DecoratedTree of D holds dom T is finite iff T in it;
existence
proof
defpred X[object] means ex T being DecoratedTree of D st $1 = T &
dom T is finite;
consider X such that
A1: x in X iff x in Trees(D) & X[x] from XBOOLE_0:sch 1;
set T = the finite DecoratedTree of D;
A2: dom T is finite;
T in Trees(D) by Def7;
then
A3: X is non empty by A1,A2;
now
let x;
assume x in X;
then x in Trees D by A1;
hence x is DecoratedTree of D;
end;
then reconsider X as DTree-set of D by A3,Def6;
take X;
let T be DecoratedTree of D;
T in Trees D by Def7;
hence dom T is finite implies T in X by A1;
assume T in X;
then ex t being DecoratedTree of D st T = t & dom t is finite by A1;
hence thesis;
end;
uniqueness
proof
let T1,T2 be DTree-set of D such that
A4: for T being DecoratedTree of D holds dom T is finite iff T in T1 and
A5: for T being DecoratedTree of D holds dom T is finite iff T in T2;
thus T1 c= T2
proof
let x be object;
assume
A6: x in T1;
then reconsider T = x as DecoratedTree of D;
dom T is finite by A4,A6;
hence thesis by A5;
end;
let x be object;
assume
A7: x in T2;
then reconsider T = x as DecoratedTree of D;
dom T is finite by A5,A7;
hence thesis by A4;
end;
end;
theorem
for D being non empty set holds FinTrees D c= Trees D
by Def7;
begin :: Functions yielding trees
definition
let IT be Function;
attr IT is Tree-yielding means
: Def9:
rng IT is constituted-Trees;
attr IT is FinTree-yielding means
: Def10:
rng IT is constituted-FinTrees;
attr IT is DTree-yielding means
rng IT is constituted-DTrees;
end;
registration
cluster empty -> Tree-yielding FinTree-yielding DTree-yielding for Function;
coherence;
end;
theorem Th22:
f is Tree-yielding iff for x st x in dom f holds f.x is Tree
proof
thus f is Tree-yielding implies for x st x in dom f holds f.x is Tree
proof
assume
A1: for x st x in rng f holds x is Tree;
let x;
assume x in dom f;
then f.x in rng f by FUNCT_1:def 3;
hence thesis by A1;
end;
assume
A2: for x st x in dom f holds f.x is Tree;
let x;
assume x in rng f;
then ex y being object st y in dom f & x = f.y by FUNCT_1:def 3;
hence thesis by A2;
end;
theorem
f is FinTree-yielding iff for x st x in dom f holds f.x is finite Tree
proof
thus f is FinTree-yielding implies for x st x in dom f holds f.x is
finite Tree
proof
assume
A1: for x st x in rng f holds x is finite Tree;
let x;
assume x in dom f;
then f.x in rng f by FUNCT_1:def 3;
hence thesis by A1;
end;
assume
A2: for x st x in dom f holds f.x is finite Tree;
let x;
assume x in rng f;
then ex y being object st y in dom f & x = f.y by FUNCT_1:def 3;
hence thesis by A2;
end;
theorem Th24:
f is DTree-yielding iff for x st x in dom f holds f.x is DecoratedTree
proof
thus f is DTree-yielding implies
for x st x in dom f holds f.x is DecoratedTree
proof
assume
A1: for x st x in rng f holds x is DecoratedTree;
let x;
assume x in dom f;
then f.x in rng f by FUNCT_1:def 3;
hence thesis by A1;
end;
assume
A2: for x st x in dom f holds f.x is DecoratedTree;
let x;
assume x in rng f;
then ex y being object st y in dom f & x = f.y by FUNCT_1:def 3;
hence thesis by A2;
end;
theorem Th25:
p is Tree-yielding & q is Tree-yielding iff p^q is Tree-yielding
proof
A1: rng (p^q) = rng p \/ rng q by FINSEQ_1:31;
thus
p is Tree-yielding & q is Tree-yielding implies p^q is Tree-yielding
by A1,Th3;
assume
A2: rng (p^q) is constituted-Trees;
hence rng p is constituted-Trees by A1,Th3;
thus rng q is constituted-Trees by A1,A2,Th3;
end;
theorem Th26:
p is FinTree-yielding & q is FinTree-yielding iff p^q is FinTree-yielding
proof
A1: rng (p^q) = rng p \/ rng q by FINSEQ_1:31;
thus p is FinTree-yielding & q is FinTree-yielding implies
p^q is FinTree-yielding
by A1,Th6;
assume
A2: rng (p^q) is constituted-FinTrees;
hence rng p is constituted-FinTrees by A1,Th6;
thus rng q is constituted-FinTrees by A1,A2,Th6;
end;
theorem Th27:
p is DTree-yielding & q is DTree-yielding iff p^q is DTree-yielding
proof
A1: rng (p^q) = rng p \/ rng q by FINSEQ_1:31;
thus
p is DTree-yielding & q is DTree-yielding implies p^q is DTree-yielding
by A1,Th9;
assume
A2: rng (p^q) is constituted-DTrees;
hence rng p is constituted-DTrees by A1,Th9;
thus rng q is constituted-DTrees by A1,A2,Th9;
end;
theorem Th28:
<*x*> is Tree-yielding iff x is Tree
proof
A1: x is Tree iff {x} is constituted-Trees by Th12;
rng <*x*> = {x} by FINSEQ_1:39;
hence thesis by A1;
end;
theorem Th29:
for x being object holds
<*x*> is FinTree-yielding iff x is finite Tree
proof let x be object;
A1: x is finite Tree iff {x} is constituted-FinTrees by Th13;
rng <*x*> = {x} by FINSEQ_1:39;
hence thesis by A1;
end;
theorem Th30:
<*x*> is DTree-yielding iff x is DecoratedTree
proof
A1: x is DecoratedTree iff {x} is constituted-DTrees by Th14;
rng <*x*> = {x} by FINSEQ_1:39;
hence thesis by A1;
end;
theorem Th31:
<*x,y*> is Tree-yielding iff x is Tree & y is Tree
proof
A1: x is Tree & y is Tree iff {x,y} is constituted-Trees by Th15;
rng <*x,y*> = {x,y} by FINSEQ_2:127;
hence thesis by A1;
end;
theorem Th32:
<*x,y*> is FinTree-yielding iff x is finite Tree & y is finite Tree
proof
A1: x is finite Tree & y is finite Tree iff {x,y} is constituted-FinTrees
by Th16;
rng <*x,y*> = {x,y} by FINSEQ_2:127;
hence thesis by A1;
end;
theorem Th33:
<*x,y*> is DTree-yielding iff x is DecoratedTree & y is DecoratedTree
proof
A1: x is DecoratedTree & y is DecoratedTree iff {x,y} is constituted-DTrees
by Th17;
rng <*x,y*> = {x,y} by FINSEQ_2:127;
hence thesis by A1;
end;
theorem Th34:
i <> 0 implies (i |-> x is Tree-yielding iff x is Tree)
proof
assume
A1: i <> 0;
i |-> x = (Seg i) --> x by FINSEQ_2:def 2;
then rng (i |-> x) = {x} by A1,FUNCOP_1:8;
then x is Tree iff rng (i |-> x) is constituted-Trees by Th12;
hence thesis;
end;
theorem Th35:
i <> 0 implies (i |-> x is FinTree-yielding iff x is finite Tree)
proof
assume
A1: i <> 0;
i |-> x = (Seg i) --> x by FINSEQ_2:def 2;
then rng (i |-> x) = {x} by A1,FUNCOP_1:8;
then x is finite Tree iff rng (i |-> x) is constituted-FinTrees by Th13;
hence thesis;
end;
theorem Th36:
i <> 0 implies (i |-> x is DTree-yielding iff x is DecoratedTree)
proof
assume
A1: i <> 0;
i |-> x = (Seg i) --> x by FINSEQ_2:def 2;
then rng (i |-> x) = {x} by A1,FUNCOP_1:8;
then x is DecoratedTree iff rng (i |-> x) is constituted-DTrees by Th14;
hence thesis;
end;
registration
cluster Tree-yielding FinTree-yielding non empty for FinSequence;
existence
proof set T = the finite Tree;
take <*T*>;
thus thesis by Th28,Th29;
end;
cluster DTree-yielding non empty for FinSequence;
existence
proof set T = the DecoratedTree;
take <*T*>;
thus thesis by Th30;
end;
end;
registration
cluster Tree-yielding FinTree-yielding non empty for Function;
existence
proof
set p = the Tree-yielding FinTree-yielding non empty FinSequence;
take p;
thus thesis;
end;
cluster DTree-yielding non empty for Function;
existence
proof set p = the DTree-yielding non empty FinSequence;
take p;
thus thesis;
end;
end;
registration
cluster FinTree-yielding -> Tree-yielding for Function;
coherence;
end;
registration
let D be constituted-Trees non empty set;
cluster -> Tree-yielding for FinSequence of D;
coherence
proof
let p be FinSequence of D;
let x;
rng p c= D;
hence thesis;
end;
end;
registration
let p,q be Tree-yielding FinSequence;
cluster p^q -> Tree-yielding;
coherence by Th25;
end;
registration
let D be constituted-FinTrees non empty set;
cluster -> FinTree-yielding for FinSequence of D;
coherence
proof
let p be FinSequence of D;
let x;
rng p c= D;
hence thesis;
end;
end;
registration
let p,q be FinTree-yielding FinSequence;
cluster p^q -> FinTree-yielding;
coherence by Th26;
end;
registration
let D be constituted-DTrees non empty set;
cluster -> DTree-yielding for FinSequence of D;
coherence
proof
let p be FinSequence of D;
let x;
rng p c= D;
hence thesis;
end;
end;
registration
let p,q be DTree-yielding FinSequence;
cluster p^q -> DTree-yielding;
coherence by Th27;
end;
registration
let T be Tree;
cluster <*T*> -> Tree-yielding non empty;
coherence by Th28;
let S be Tree;
cluster <*T,S*> -> Tree-yielding non empty;
coherence by Th31;
end;
registration
let n be Nat, T be Tree;
cluster n |-> T -> Tree-yielding;
coherence
proof 0 |-> T = {};
hence thesis by Th34;
end;
end;
registration
let T be finite Tree;
cluster <*T*> -> FinTree-yielding;
coherence by Th29;
let S be finite Tree;
cluster <*T,S*> -> FinTree-yielding;
coherence by Th32;
end;
registration
let n be Nat, T be finite Tree;
cluster n |-> T -> FinTree-yielding;
coherence
proof 0 |-> T = {};
hence thesis by Th35;
end;
end;
registration
let T be DecoratedTree;
cluster <*T*> -> DTree-yielding non empty;
coherence by Th30;
let S be DecoratedTree;
cluster <*T,S*> -> DTree-yielding non empty;
coherence by Th33;
end;
registration
let n be Nat, T be DecoratedTree;
cluster n |-> T -> DTree-yielding;
coherence
proof 0 |-> T = {};
hence thesis by Th36;
end;
end;
registration
cluster DTree-yielding -> Function-yielding for Function;
coherence
proof let f be Function;
assume
A1: rng f is constituted-DTrees;
let x be object;
assume x in dom f;
then f.x in rng f by FUNCT_1:3;
then f.x is DecoratedTree by A1;
hence f.x is Function;
end;
end;
theorem Th37:
for f being DTree-yielding Function holds dom doms f = dom f &
doms f is Tree-yielding
proof
let f be DTree-yielding Function;
A1: dom doms f = dom f by FUNCT_6:def 2;
hence dom doms f c= dom f;
thus dom f c= dom doms f by A1;
now
let x;
assume x in dom doms f;
then
A2: x in dom f by A1;
then reconsider g = f.x as DecoratedTree by Th24;
(doms f).x = dom g by A2,FUNCT_6:22;
hence (doms f).x is Tree;
end;
hence thesis by Th22;
end;
registration
let p be DTree-yielding FinSequence;
cluster doms p -> Tree-yielding FinSequence-like;
coherence
proof
A1: dom doms p = dom p by Th37;
Seg len p = dom p by FINSEQ_1:def 3;
hence thesis by A1,Th37,FINSEQ_1:def 2;
end;
end;
theorem
for p being DTree-yielding FinSequence holds len doms p = len p
proof
let p be DTree-yielding FinSequence;
A1: dom p = dom doms p by Th37;
Seg len p = dom p by FINSEQ_1:def 3;
hence thesis by A1,FINSEQ_1:def 3;
end;
Lm2: for D being non empty set, T being DecoratedTree of D holds
T is Function of dom T, D
proof
let D be non empty set, T be DecoratedTree of D;
rng T c= D;
hence thesis by FUNCT_2:def 1,RELSET_1:4;
end;
begin :: Trees decorated by Cartesian product and structure of substitution
definition
let D,E be non empty set;
mode DecoratedTree of D,E is DecoratedTree of [:D,E:];
mode DTree-set of D,E is DTree-set of [:D,E:];
end;
registration
let T1,T2 be DecoratedTree;
cluster <:T1,T2:> -> DecoratedTree-like;
coherence
proof dom <:T1,T2:> = dom T1 /\ dom T2 by FUNCT_3:def 7;
hence thesis;
end;
end;
registration
let D1,D2 be non empty set;
let T1 be DecoratedTree of D1;
let T2 be DecoratedTree of D2;
cluster <:T1,T2:> -> [:D1,D2:]-valued;
coherence
proof
A1: rng <:T1,T2:> c= [:rng T1, rng T2:] by FUNCT_3:51;
[:rng T1, rng T2:] c= [:D1,D2:] by ZFMISC_1:96;
then rng <:T1,T2:> c= [:D1,D2:] by A1;
hence thesis by RELAT_1:def 19;
end;
end;
registration
let D,E be non empty set;
let T be DecoratedTree of D;
let f be Function of D,E;
cluster f*T -> DecoratedTree-like;
coherence
proof reconsider g = T as Function of dom T, D by Lm2;
reconsider fg = f*g as Function of dom T, E;
rng fg c= E;
hence thesis;
end;
end;
definition
let D1,D2 be non empty set;
redefine func pr1(D1,D2) -> Function of [:D1,D2:], D1;
coherence
proof
thus pr1(D1,D2) is Function of [:D1,D2:], D1;
end;
redefine func pr2(D1,D2) -> Function of [:D1,D2:], D2;
coherence
proof
thus pr2(D1,D2) is Function of [:D1,D2:], D2;
end;
end;
definition
let D1,D2 be non empty set, T be DecoratedTree of D1,D2;
func T`1 -> DecoratedTree of D1 equals
pr1(D1,D2)*T;
correctness;
func T`2 -> DecoratedTree of D2 equals
pr2(D1,D2)*T;
correctness;
end;
theorem Th39:
for D1,D2 being non empty set, T being DecoratedTree of D1,D2,
t being Element of dom T holds (T.t)`1 = T`1.t & T`2.t = (T.t)`2
proof
let D1,D2 be non empty set, T be DecoratedTree of D1,D2;
let t be Element of dom T;
A1: dom pr1(D1,D2) = [:D1,D2:] by FUNCT_2:def 1;
A2: dom pr2(D1,D2) = [:D1,D2:] by FUNCT_2:def 1;
A3: rng T c= [:D1,D2:];
then
A4: dom T`1 = dom T by A1,RELAT_1:27;
A5: dom T`2 = dom T by A2,A3,RELAT_1:27;
A6: T.t = [(T.t)`1,(T.t)`2] by MCART_1:21;
then
A7: T`1.t = pr1(D1,D2).((T.t)`1,(T.t)`2) by A4,FUNCT_1:12;
T`2.t = pr2(D1,D2).((T.t)`1,(T.t)`2) by A5,A6,FUNCT_1:12;
hence thesis by A7,FUNCT_3:def 4,def 5;
end;
theorem
for D1,D2 being non empty set, T being DecoratedTree of D1,D2 holds
<:T`1,T`2:> = T
proof
let D1,D2 be non empty set, T be DecoratedTree of D1,D2;
A1: dom pr1(D1,D2) = [:D1,D2:] by FUNCT_2:def 1;
A2: dom pr2(D1,D2) = [:D1,D2:] by FUNCT_2:def 1;
A3: rng T c= [:D1,D2:];
then
A4: dom T`1 = dom T by A1,RELAT_1:27;
A5: dom T`2 = dom T by A2,A3,RELAT_1:27;
then
A6: dom <:T`1,T`2:> = dom T by A4,FUNCT_3:50;
now
let x be object;
assume x in dom T;
then reconsider t = x as Element of dom T;
thus <:T`1,T`2:>.x = [T`1.t,T`2.t] by A4,A5,FUNCT_3:49
.= [(T.t)`1,T`2.t] by Th39
.= [(T.t)`1,(T.t)`2] by Th39
.= T.x by MCART_1:21;
end;
hence thesis by A6;
end;
registration
let T be finite Tree;
cluster Leaves T -> finite non empty;
coherence
proof
A1: T <> {};
defpred X[object,object] means
ex t1,t2 being Element of T st $1 = t1 & $2 = t2 & t2 is_a_prefix_of t1;
A2: for x,y st X[x,y] & X[y,x] holds x = y by XBOOLE_0:def 10;
A3: for x,y,z st X[x,y] & X[y,z] holds X[x,z] by XBOOLE_1:1;
consider x such that
A4: x in T & for y st y in T & y <> x holds not X[y,x]
from CARD_2:sch 1(A1,A2,A3);
reconsider x as Element of T by A4;
now
let p be FinSequence of NAT;
assume p in T;
then reconsider t1 = p as Element of T;
thus not x is_a_proper_prefix_of p
proof
assume x is_a_prefix_of p;
then x = t1 by A4;
hence thesis;
end;
end;
hence thesis by TREES_1:def 5;
end;
end;
definition
let T be Tree, S be non empty Subset of T;
redefine mode Element of S -> Element of T;
coherence
proof
let t be Element of S;
thus thesis;
end;
end;
definition
let T be finite Tree;
redefine mode Leaf of T -> Element of Leaves T;
coherence by TREES_1:def 7;
end;
definition
let T be finite Tree;
mode T-Substitution of T -> Tree means
:
Def14: for t being Element of it holds t in T or
ex l being Leaf of T st l is_a_proper_prefix_of t;
existence
proof
take T;
thus thesis;
end;
end;
definition
let T be finite Tree, t be Leaf of T, S be Tree;
redefine func T with-replacement (t,S) -> T-Substitution of T;
coherence
proof
let s be Element of T with-replacement (t,S);
assume
A1: not s in T;
then consider t1 being FinSequence of NAT such that
t1 in S and
A2: s = t^t1 by TREES_1:def 9;
take t;
t^{} = t by FINSEQ_1:34;
then t1 <> {} by A1,A2;
hence thesis by A2,TREES_1:10;
end;
end;
registration
let T be finite Tree;
cluster finite for T-Substitution of T;
existence
proof
for t being Element of T holds t in T or
ex l being Leaf of T st l is_a_proper_prefix_of t;
then T is T-Substitution of T by Def14;
hence thesis;
end;
end;
definition
let n;
mode T-Substitution of n is T-Substitution of elementary_tree n;
end;
theorem
for T being Tree holds T is T-Substitution of 0
proof
let T be Tree;
let t be Element of T;
assume
A1: not t in elementary_tree 0;
set l = the Leaf of elementary_tree 0;
take l;
A2: l = {} by TARSKI:def 1,TREES_1:29;
A3: t <> {} by A1,TARSKI:def 1,TREES_1:29;
{} is_a_prefix_of t;
hence thesis by A2,A3;
end;
theorem
for T1, T2 being Tree st T1-level 1 c= T2-level 1 &
for n being Element of NAT st <*n*> in T1 holds T1|<*n*> = T2|<*n*>
holds T1 c= T2
proof
let T1, T2 be Tree;
assume that
A1: T1-level 1 c= T2-level 1 and
A2: for n being Element of NAT st <*n*> in T1 holds T1|<*n*> = T2|<*n*>;
let x be object;
assume x in T1;
then reconsider t = x as Element of T1;
now
assume t <> {};
then consider p being FinSequence of NAT,
n being Element of NAT such that
A3: t = <*n*>^p by FINSEQ_2:130;
A4: len <*n*> = 1 by FINSEQ_1:39;
reconsider n as Element of NAT;
reconsider q = <*n*> as Element of T1 by A3,TREES_1:21;
A5: q in T1-level 1 by A4;
then
A6: q in T2-level 1 by A1;
A7: p in T1|q by A3,TREES_1:def 6;
T1|<*n*> = T2|<*n*> by A2,A5;
hence t in T2 by A3,A6,A7,TREES_1:def 6;
end;
hence thesis by TREES_1:22;
end;
Lm3: n < len p implies n+1 in dom p & p.(n+1) in rng p
proof
assume
A1: n < len p;
A2: n+1 >= 0+1 by NAT_1:13;
n+1 <= len p by A1,NAT_1:13;
then n+1 in dom p by A2,FINSEQ_3:25;
hence thesis by FUNCT_1:def 3;
end;
begin :: Joining of trees
theorem
for T,T9 being Tree, p being FinSequence of NAT st
p in Leaves T holds T c= T with-replacement (p,T9)
proof
let T,T9 be Tree, p be FinSequence of NAT such that
A1: p in Leaves T;
let x be object;
assume x in T;
then reconsider t = x as Element of T;
not p is_a_proper_prefix_of t by A1,TREES_1:def 5;
hence thesis by A1,TREES_1:def 9;
end;
theorem
for T,T9 being DecoratedTree, p being Element of dom T holds
(T with-replacement (p,T9)).p = T9.{}
proof
let T,T9 be DecoratedTree, p be Element of dom T;
p in dom T with-replacement (p,dom T9) by TREES_1:def 9;
then
A1: ex r being FinSequence of NAT st ( r in dom T9)&( p = p^r)&(
(T with-replacement (p,T9)).p = T9.r) by TREES_2:def 11;
p = p^{} by FINSEQ_1:34;
hence thesis by A1,FINSEQ_1:33;
end;
theorem
for T,T9 being DecoratedTree, p,q being Element of dom T st
not p is_a_prefix_of q holds (T with-replacement (p,T9)).q = T.q
proof
let T,T9 be DecoratedTree, p,q be Element of dom T;
assume
A1: not p is_a_prefix_of q;
then
A2: q in dom T with-replacement(p,dom T9) by TREES_2:7;
not ex r being FinSequence of NAT st r in dom T9 & q = p^r & (T
with-replacement(p,T9)).q = T9.r by A1,TREES_1:1;
hence thesis by A2,TREES_2:def 11;
end;
theorem
for T,T9 being DecoratedTree, p being Element of dom T,
q being Element of dom T9 holds (T with-replacement (p,T9)).(p^q) = T9.q
proof
let T,T9 be DecoratedTree;
let p be Element of dom T, q be Element of dom T9;
A1: p is_a_prefix_of p^q by TREES_1:1;
p^q in dom T with-replacement(p,dom T9) by TREES_1:def 9;
then ex r being FinSequence of NAT st ( r in dom T9)&( p^q = p^r
)&( (T with-replacement (p,T9)).(p^q) = T9.r) by A1,TREES_2:def 11;
hence thesis by FINSEQ_1:33;
end;
registration
let T1,T2 be Tree;
cluster T1 \/ T2 -> non empty Tree-like;
coherence;
end;
theorem Th47:
for T1,T2 being Tree, p being Element of T1 \/ T2 holds
(p in T1 & p in T2 implies (T1 \/ T2)|p = (T1|p) \/ (T2|p)) &
(not p in T1 implies (T1 \/ T2)|p = T2|p) &
(not p in T2 implies (T1 \/ T2)|p = T1|p)
proof
let T1,T2 be Tree, p be Element of T1 \/ T2;
thus p in T1 & p in T2 implies (T1 \/ T2)|p = (T1|p) \/ (T2|p)
proof
assume that
A1: p in T1 and
A2: p in T2;
let q be FinSequence of NAT;
thus q in (T1 \/ T2)|p implies q in (T1|p) \/ (T2|p)
proof
assume q in (T1 \/ T2)|p;
then p^q in T1 \/ T2 by TREES_1:def 6;
then p^q in T1 or p^q in T2 by XBOOLE_0:def 3;
then q in T1|p or q in T2|p by A1,A2,TREES_1:def 6;
hence thesis by XBOOLE_0:def 3;
end;
assume q in (T1|p) \/ (T2|p);
then q in T1|p or q in T2|p by XBOOLE_0:def 3;
then p^q in T1 or p^q in T2 by A1,A2,TREES_1:def 6;
then p^q in T1 \/ T2 by XBOOLE_0:def 3;
hence thesis by TREES_1:def 6;
end;
for T1,T2 being Tree, p being Element of T1 \/ T2 st
not p in T1 holds (T1 \/ T2)|p = T2|p
proof
let T1, T2 be Tree;
let p be Element of T1 \/ T2;
assume
A3: not p in T1;
then
A4: p in T2 by XBOOLE_0:def 3;
let q be FinSequence of NAT;
thus q in (T1 \/ T2)|p implies q in T2|p
proof
assume q in (T1 \/ T2)|p;
then p^q in T1 \/ T2 by TREES_1:def 6;
then p^q in T1 or p^q in T2 by XBOOLE_0:def 3;
hence thesis by A3,A4,TREES_1:21,def 6;
end;
assume q in T2|p;
then p^q in T2 by A4,TREES_1:def 6;
then p^q in T1 \/ T2 by XBOOLE_0:def 3;
hence thesis by TREES_1:def 6;
end;
hence thesis;
end;
definition
let p such that
A1: p is Tree-yielding;
func tree p -> Tree means
:Def15:
x in it iff x = {} or ex n,q st n < len p & q in p.(n+1) & x = <*n*>^q;
existence
proof
defpred X[object] means
($1 = {} or ex n,q st n < len p & q in p.(n+1) & $1 = <*n*>^q);
consider T being set such that
A2: for y being object holds y in T iff y in NAT* & X[y] from XBOOLE_0:sch 1;
<*>NAT in NAT* by FINSEQ_1:def 11;
then reconsider T as non empty set by A2;
A3: rng p is constituted-Trees by A1;
then
A4: for n st n < len p holds p.(n+1) is Tree by Lm3;
T is Tree-like
proof
thus T c= NAT*
by A2;
thus w in T implies ProperPrefixes w c= T
proof
assume
A5: w in T;
now
assume w <> {};
then consider n,q such that
A6: n < len p and
A7: q in p.(n+1) and
A8: w = <*n*>^q by A2,A5;
reconsider q as FinSequence of NAT by A8,FINSEQ_1:36;
thus thesis
proof
let x be object;
assume x in ProperPrefixes w;
then consider r such that
A9: x = r and
A10: r is_a_proper_prefix_of w by TREES_1:def 2;
r is_a_prefix_of w by A10;
then consider k such that
A11: r = w|Seg k;
rng r = w.:Seg k by A11,RELAT_1:115;
then reconsider r as FinSequence of NAT by FINSEQ_1:def 4;
A12: r in NAT* by FINSEQ_1:def 11;
now
assume r <> {};
then consider r1 being FinSequence of NAT,
i being Element of NAT such that
A13: r = <*i*>^r1 by FINSEQ_2:130;
A14: dom <*i*> = Seg 1 by FINSEQ_1:38;
A15: 1 in Seg 1 by FINSEQ_1:2,TARSKI:def 1;
A16: Seg 1 c= dom r by A13,A14,FINSEQ_1:26;
A17: r.1 = i by A13,FINSEQ_1:41;
A18: w.1 = n by A8,FINSEQ_1:41;
A19: r.1 = w.1 by A11,A15,A16,FUNCT_1:47;
then
A20: r1 is_a_proper_prefix_of q by A8,A10,A13,A17,A18,TREES_1:49;
A21: p.(n+1) is Tree by A4,A6;
r1 is_a_prefix_of q by A20;
then r1 in p.(n+1) by A7,A21,TREES_1:20;
hence thesis by A2,A6,A9,A12,A13,A17,A18,A19;
end;
hence thesis by A2,A9,A12;
end;
end;
hence thesis by TREES_1:15;
end;
let w,k,n such that
A22: w^<*k*> in T and
A23: n <= k;
A24: now
assume
A25: w = {};
then <*k*> in T by A22,FINSEQ_1:34;
then consider m being Nat,q such that
A26: m < len p and q in p.(m+1) and
A27: <*k*> = <*m*>^q by A2;
A28: <*k*>.1 = k by FINSEQ_1:def 8;
(<*m*>^q).1 = m by FINSEQ_1:41;
then
A29: n < len p by A23,A26,A27,A28,XXREAL_0:2;
then p.(n+1) in rng p by Lm3;
then
A30: {} in p.(n+1) by A3,TREES_1:22;
A31: <*n*>^{} = <*n*> by FINSEQ_1:34;
A32: {}^<*n*> = <*n*> by FINSEQ_1:34;
reconsider n as Element of NAT by ORDINAL1:def 12;
<*n*> in NAT* by FINSEQ_1:def 11;
hence thesis by A2,A25,A29,A30,A31,A32;
end;
now
assume w <> {};
then consider q being FinSequence of NAT,
m being Element of NAT such that
A33: w = <*m*>^q by FINSEQ_2:130;
consider m9 being Nat,r such that
A34: m9 < len p and
A35: r in p.(m9+1) and
A36: w^<*k*> = <*m9*>^r by A2,A22;
reconsider n as Element of NAT by ORDINAL1:def 12;
A37: w^<*k*> = <*m*>^(q^<*k*>) by A33,FINSEQ_1:32;
A38: w^<*n*> = <*m*>^(q^<*n*>) by A33,FINSEQ_1:32;
A39: (w^<*k*>).1 = m by A37,FINSEQ_1:41;
A40: (w^<*k*>).1 = m9 by A36,FINSEQ_1:41;
A41: <*m*>^(q^<*n*>) in NAT* by FINSEQ_1:def 11;
A42: r = q^<*k*> by A36,A37,A39,A40,FINSEQ_1:33;
p.(m+1) in rng p by A34,A39,A40,Lm3;
then q^<*n*> in p.(m+1) by A3,A23,A35,A39,A40,A42,TREES_1:def 3;
hence thesis by A2,A34,A38,A39,A40,A41;
end;
hence thesis by A24;
end;
then reconsider T as Tree;
take T;
let x;
thus x in T implies x = {} or
ex n,q st n < len p & q in p.(n+1) & x = <*n*>^q by A2;
assume
A43: x = {} or ex n,q st n < len p & q in p.(n+1) & x = <*n*>^q;
A44: now
given n,q such that
A45: n < len p and
A46: q in p.(n+1) and
A47: x = <*n*>^q;
reconsider T1 = p.(n+1) as Tree by A4,A45;
reconsider q as Element of T1 by A46;
reconsider n as Element of NAT by ORDINAL1:def 12;
<*n*>^q in NAT* by FINSEQ_1:def 11;
hence x in NAT* by A47;
end;
{} in NAT* by FINSEQ_1:49;
hence thesis by A2,A43,A44;
end;
uniqueness
proof
defpred X[object] means
$1 = {} or ex n,q st n < len p & q in p.(n+1) & $1 = <*n*>^q;
let T1,T2 be Tree such that
A48: x in T1 iff X[x] and
A49: x in T2 iff X[x];
thus thesis from XBOOLE_0:sch 2(A48,A49);
end;
end;
definition
let T be Tree;
func ^T -> Tree equals
tree<*T*>;
correctness;
end;
definition
let T1,T2 be Tree;
func tree(T1,T2) -> Tree equals
tree(<*T1,T2*>);
correctness;
end;
theorem
p is Tree-yielding implies (<*n*>^q in tree(p) iff n < len p & q in p.(n+ 1
) )
proof
assume
A1: p is Tree-yielding;
thus <*n*>^q in tree(p) implies n < len p & q in p.(n+1)
proof
assume <*n*>^q in tree(p);
then consider k,r such that
A2: k < len p and
A3: r in p.(k+1) and
A4: <*n*>^q = <*k*>^r by A1,Def15;
A5: (<*n*>^q).1 = n by FINSEQ_1:41;
(<*k*>^r).1 = k by FINSEQ_1:41;
hence thesis by A2,A3,A4,A5,FINSEQ_1:33;
end;
thus thesis by A1,Def15;
end;
theorem Th49:
p is Tree-yielding implies tree(p)-level 1 = {<*n*>: n < len p} &
for n being Element of NAT st n < len p holds (tree(p))|<*n*> = p.(n+1)
proof
set T = tree(p);
assume
A1: p is Tree-yielding;
then
A2: rng p is constituted-Trees;
thus T-level 1 = {<*n*>: n < len p}
proof
thus T-level 1 c= {<*n*>: n < len p}
proof
let x be object;
assume x in T-level 1;
then consider t being Element of T such that
A3: x = t and
A4: len t = 1;
A5: t = <*t.1*> by A4,FINSEQ_1:40;
then consider n, q such that
A6: n < len p and q in p.(n+1) and
A7: t = <*n*>^q by A1,Def15;
t = <*n*> by A5,A7,FINSEQ_1:88;
hence thesis by A3,A6;
end;
let x be object;
assume x in {<*n*>: n < len p};
then consider n such that
A8: x = <*n*> and
A9: n < len p;
p.(n+1) in rng p by A9,Lm3;
then
A10: {} in p.(n+1) by A2,TREES_1:22;
<*n*>^{} = <*n*> by FINSEQ_1:34;
then reconsider t = <*n*> as Element of T by A1,A9,A10,Def15;
len t = 1 by FINSEQ_1:39;
hence thesis by A8;
end;
let n be Element of NAT;
assume
A11: n < len p;
then p.(n+1) in rng p by Lm3;
then reconsider S = p.(n+1) as Tree by A2;
A12: {} in S by TREES_1:22;
<*n*>^{} = <*n*> by FINSEQ_1:34;
then
A13: <*n*> in T by A1,A11,A12,Def15;
T|<*n*> = S
proof
let r be FinSequence of NAT;
thus r in T|<*n*> implies r in S
proof
assume r in T|<*n*>;
then <*n*>^r in T by A13,TREES_1:def 6;
then consider m being Nat, q such that
m < len p and
A14: q in p.(m+1) and
A15: <*n*>^r = <*m*>^q by A1,Def15;
A16: (<*n*>^r).1 = n by FINSEQ_1:41;
(<*m*>^q).1 = m by FINSEQ_1:41;
hence thesis by A14,A15,A16,FINSEQ_1:33;
end;
assume r in S;
then
A17: <*n*>^r in T by A1,A11,Def15;
then <*n*> in T by TREES_1:21;
hence thesis by A17,TREES_1:def 6;
end;
hence thesis;
end;
theorem
for p,q being Tree-yielding FinSequence st tree(p) = tree(q) holds p = q
proof
let p,q be Tree-yielding FinSequence such that
A1: tree(p) = tree(q);
A2: tree(p)-level 1 = {<*n*>: n < len p} by Th49;
A3: tree(q)-level 1 = {<*k*>: k < len q} by Th49;
A4: now
let n1,n2 be Element of NAT;
assume {<*i*>: i < n1} = {<*k*>: k < n2} & n1 < n2;
then <*n1*> in {<*i*>: i < n1};
then
A5: ex i st ( <*n1*> = <*i*>)&( i < n1);
<*n1*>.1 = n1 by FINSEQ_1:40;
hence contradiction by A5,FINSEQ_1:40;
end;
then
A6: not len p < len q by A1,A2,A3;
A7: not len p > len q by A1,A2,A3,A4;
then
A8: len p = len q by A6,XXREAL_0:1;
now
let i be Nat;
assume that
A9: i >= 1 and
A10: i <= len p;
consider k be Nat such that
A11: i = 1+k by A9,NAT_1:10;
reconsider k as Element of NAT by ORDINAL1:def 12;
A12: k < len p by A10,A11,NAT_1:13;
then p.i = (tree(p))|<*k*> by A11,Th49;
hence p.i = q.i by A1,A8,A11,A12,Th49;
end;
hence thesis by A6,A7,FINSEQ_1:14,XXREAL_0:1;
end;
theorem Th51:
for p1,p2 being Tree-yielding FinSequence, T being Tree holds
p in T iff <*len p1*>^p in tree(p1^<*T*>^p2)
proof
let p1,p2 be Tree-yielding FinSequence, T be Tree;
A1: len (p1^<*T*>^p2) = len (p1^<*T*>) + len p2 by FINSEQ_1:22;
A2: len (p1^<*T*>) = len p1 + len <*T*> by FINSEQ_1:22;
A3: len <*T*> = 1 by FINSEQ_1:40;
A4: len p1+1+len p2 = (len p1+len p2)+1;
len p1 <= len p1+len p2 by NAT_1:11;
then
A5: len p1 < len (p1^<*T*>^p2) by A1,A2,A3,A4,NAT_1:13;
len p1+1 >= 1 by NAT_1:11;
then len p1+1 in dom (p1^<*T*>) by A2,A3,FINSEQ_3:25;
then
A6: (p1^<*T*>^p2).(len p1+1) = (p1^<*T*>).(len p1+1) by FINSEQ_1:def 7
.= T by FINSEQ_1:42;
hence p in T implies <*len p1*>^p in tree(p1^<*T*>^p2) by A5,Def15;
assume <*len p1*>^p in tree(p1^<*T*>^p2);
then consider n,q such that
n < len (p1^<*T*>^p2) and
A7: q in (p1^<*T*>^p2).(n+1) and
A8: <*len p1*>^p = <*n*>^q by Def15;
A9: (<*len p1*>^p).1 = len p1 by FINSEQ_1:41;
(<*n*>^q).1 = n by FINSEQ_1:41;
hence thesis by A6,A7,A8,A9,FINSEQ_1:33;
end;
theorem Th52:
tree({}) = elementary_tree 0
proof
let p be FinSequence of NAT;
thus p in tree({}) implies p in elementary_tree 0
proof
assume p in tree({});
then
A1: p = {} or ex n,q st n < len {} & q in {} .(n+1) & p = <*n*>^q by Def15;
assume not thesis;
hence contradiction by A1,TARSKI:def 1,TREES_1:29;
end;
assume p in elementary_tree 0;
then p = {} by TARSKI:def 1,TREES_1:29;
hence thesis by Def15;
end;
theorem Th53:
p is Tree-yielding implies elementary_tree len p c= tree(p)
proof
assume
A1: p is Tree-yielding;
then
A2: rng p is constituted-Trees;
let q be object;
assume q in elementary_tree len p;
then q in {<*n*>: n < len p} or q in {{}} by XBOOLE_0:def 3;
then
A3: (ex n st q = <*n*> & n < len p) or q = {} by TARSKI:def 1;
assume
A4: not thesis;
then consider n such that
A5: q = <*n*> and
A6: n < len p by A3,TREES_1:22;
p.(n+1) in rng p by A6,Lm3;
then reconsider t = p.(n+1) as Tree by A2;
A7: {} in t by TREES_1:22;
<*n*>^{} = q by A5,FINSEQ_1:34;
hence thesis by A1,A4,A6,A7,Def15;
end;
theorem Th54:
elementary_tree i = tree(i|->elementary_tree 0)
proof
set p = i |-> elementary_tree 0;
let q be FinSequence of NAT;
A1: len p = i by CARD_1:def 7;
then elementary_tree i c= tree(p) by Th53;
hence q in elementary_tree i implies q in tree(p);
assume q in tree(p);
then
A2: q = {} or
ex n,r st n < len p & r in p.(n+1) & q = <*n*>^r by Def15;
now
given n, r such that
A3: n < len p and
A4: r in p.(n+1) and
A5: q = <*n*>^r;
p = (Seg i) --> elementary_tree 0 by FINSEQ_2:def 2;
then
A6: rng p c= {elementary_tree 0} by FUNCOP_1:13;
p.(n+1) in rng p by A3,Lm3;
then p.(n+1) = elementary_tree 0 by A6,TARSKI:def 1;
then r = {} by A4,TARSKI:def 1,TREES_1:29;
then q = <*n*> by A5,FINSEQ_1:34;
hence thesis by A1,A3,TREES_1:28;
end;
hence thesis by A2,TREES_1:22;
end;
theorem Th55:
for T being Tree, p being Tree-yielding FinSequence holds
tree(p^<*T*>) = (tree(p) \/ elementary_tree (len p + 1))
with-replacement (<*len p*>, T)
proof
let T be Tree, p be Tree-yielding FinSequence;
set W1 = elementary_tree (len p + 1), W2 = tree(p) \/ W1,
W = W2 with-replacement (<*len p*>, T);
len <*T*> = 1 by FINSEQ_1:40;
then
A1: len (p^<*T*>) = len p + 1 by FINSEQ_1:22;
len p < len p+1 by NAT_1:13;
then <*len p*> in W1 by TREES_1:28;
then
A2: <*len p*> in W2 by XBOOLE_0:def 3;
let q be FinSequence of NAT;
thus q in tree(p^<*T*>) implies q in W
proof
assume q in tree(p^<*T*>);
then
A3: q = {} or ex n,r st n < len (p^<*T*>) & r in (p^<*T*>).(n+1) &
q = <*n*>^r by Def15;
now
given n, r such that
A4: n < len (p^<*T*>) and
A5: r in (p^<*T*>).(n+1) and
A6: q = <*n*>^r;
reconsider r as FinSequence of NAT by A6,FINSEQ_1:36;
A7: n <= len p by A1,A4,NAT_1:13;
A8: now
assume
A9: n < len p;
then n+1 in dom p by Lm3;
then (p^<*T*>).(n+1) = p.(n+1) by FINSEQ_1:def 7;
then q in tree(p) by A5,A6,A9,Def15;
then
A10: q in W2 by XBOOLE_0:def 3;
not <*len p*> is_a_prefix_of <*n*>^r by A9,TREES_1:50;
then not <*len p*> is_a_proper_prefix_of <*n*>^r;
hence thesis by A2,A6,A10,TREES_1:def 9;
end;
now
assume
A11: n = len p;
then
A12: (p^<*T*>).(n+1) = T by FINSEQ_1:42;
r = r;
hence thesis by A2,A5,A6,A11,A12,TREES_1:def 9;
end;
hence thesis by A7,A8,XXREAL_0:1;
end;
hence thesis by A3,TREES_1:22;
end;
assume
A13: q in W;
assume
A14: not thesis;
A15: now
given r being FinSequence of NAT such that
A16: r in T and
A17: q = <*len p*>^r;
A18: len p < len p+1 by NAT_1:13;
(p^<*T*>).(len p+1) = T by FINSEQ_1:42;
hence thesis by A1,A14,A16,A17,A18,Def15;
end;
now
assume q in W2;
then
A19: q in tree(p) or q in W1 by XBOOLE_0:def 3;
A20: now
assume q in tree(p);
then q = {} & {} in tree(p^<*T*>) or
ex n,r st n < len p & r in p.(n+1) & q = <*n*>^r by Def15;
then consider n,r such that
A21: n < len p and
A22: r in p.(n+1) and
A23: q = <*n*>^r by A14;
n+1 in dom p by A21,Lm3;
then
A24: p.(n+1) = (p^<*T*>).(n+1) by FINSEQ_1:def 7;
n < len (p^<*T*>) by A1,A21,NAT_1:13;
hence thesis by A14,A22,A23,A24,Def15;
end;
now
assume
A25: not q in tree(p);
then q = {} or ex n st n < len p+1 & q = <*n*> by A19,TREES_1:30;
then consider n such that
A26: n < len p+1 and
A27: q = <*n*> by A14,TREES_1:22;
A28: now
assume n < len p;
then
A29: p.(n+1) in rng p by Lm3;
rng p is constituted-Trees by Def9;
hence {} in p.(n+1) by A29,TREES_1:22;
end;
A30: q = <*n*>^{} by A27,FINSEQ_1:34;
then
A31: n >= len p by A25,A28,Def15;
n <= len p by A26,NAT_1:13;
then
A32: len p = n by A31,XXREAL_0:1;
A33: n < n+1 by NAT_1:13;
A34: (p^<*T*>).(len p+1) = T by FINSEQ_1:42;
{} in T by TREES_1:22;
hence thesis by A1,A14,A30,A32,A33,A34,Def15;
end;
hence thesis by A20;
end;
hence thesis by A2,A13,A15,TREES_1:def 9;
end;
theorem
for p being Tree-yielding FinSequence holds
tree(p^<*elementary_tree 0*>) = tree(p) \/ elementary_tree (len p + 1)
proof
let p be Tree-yielding FinSequence;
len p < len p+1 by NAT_1:13;
then
A1: <*len p*> in elementary_tree (len p + 1) by TREES_1:28;
then
A2: <*len p*> in tree(p) \/ elementary_tree (len p + 1) by XBOOLE_0:def 3;
{} in (elementary_tree (len p+1))|<*len p*> by TREES_1:22;
then
A3: elementary_tree 0 c= (elementary_tree (len p+1))|<*len p*>
by TREES_1:29,ZFMISC_1:31;
A4: (elementary_tree (len p+1))|<*len p*> c= elementary_tree 0
proof
let x be object;
assume x in (elementary_tree (len p+1))|<*len p*>;
then reconsider q = x as Element of (elementary_tree (len p+1))|<*len p*>;
<*len p*>^q in elementary_tree (len p+1) by A1,TREES_1:def 6;
then consider n such that
n < len p+1 and
A5: <*len p*>^q = <*n*> by TREES_1:30;
A6: len <*n*> = 1 by FINSEQ_1:40;
len <*len p*> = 1 by FINSEQ_1:40;
then 1+len q = 1+0 by A5,A6,FINSEQ_1:22;
then x = {};
hence thesis by TREES_1:22;
end;
now
let n,r;
assume <*len p*> = <*n*>^r;
then n = <*len p*>.1 by FINSEQ_1:41
.= len p by FINSEQ_1:40;
hence not n < len p;
end;
then not ex n,q st n < len p & q in p.(n+1) & <*len p*> = <*n*>^q;
then not <*len p*> in tree(p) by Def15;
then
A7: (tree(p) \/ elementary_tree (len p + 1))|<*len p*>
= (elementary_tree (len p+1))|<*len p*> by A2,Th47
.= elementary_tree 0 by A3,A4;
thus tree(p^<*elementary_tree 0*>)
= (tree(p) \/ elementary_tree (len p + 1)) with-replacement(<*len p*>,
elementary_tree 0) by Th55
.= tree(p) \/ elementary_tree (len p + 1) by A2,A7,TREES_2:6;
end;
theorem Th57:
for p, q being Tree-yielding FinSequence for T1,T2 be Tree holds
tree(p^<*T1*>^q) = tree(p^<*T2*>^q) with-replacement(<*len p*>,T1)
proof
let p, q be Tree-yielding FinSequence, T1, T2 be Tree;
set w1 = p^<*T1*>^q, W1 = tree(w1), w2 = p^<*T2*>^q, W2 = tree(w2),
W = W2 with-replacement(<*len p*>, T1);
A1: len <*T1*> = 1 by FINSEQ_1:40;
A2: len <*T2*> = 1 by FINSEQ_1:40;
A3: len (p^<*T1*>) = len p + 1 by A1,FINSEQ_1:22;
A4: len w1 = len (p^<*T1*>)+len q by FINSEQ_1:22;
A5: len (p^<*T2*>) = len p + 1 by A2,FINSEQ_1:22;
A6: len w2 = len (p^<*T2*>)+len q by FINSEQ_1:22;
len p+1 <= len p+1+len q by NAT_1:11;
then
A7: len p < len p+1+len q by NAT_1:13;
then
A8: w2.(len p+1) in rng w2 by A5,A6,Lm3;
rng w2 is constituted-Trees by Def9;
then
A9: {} in w2.(len p+1) by A8,TREES_1:22;
<*len p*>^{} = <*len p*> by FINSEQ_1:34;
then
A10: <*len p*> in W2 by A5,A6,A7,A9,Def15;
let r be FinSequence of NAT;
thus r in W1 implies r in W
proof
assume r in W1;
then
A11: r = {} or ex n,s st n < len w1 & s in w1.(n+1) & r = <*n*>^s by Def15;
now
given n, s such that
A12: n < len w1 and
A13: s in w1.(n+1) and
A14: r = <*n*>^s;
reconsider s as FinSequence of NAT by A14,FINSEQ_1:36;
A15: n <= len p or n >= len p+1 by NAT_1:13;
A16: now
assume
A17: n < len p;
then
A18: n+1 in dom p by Lm3;
A19: dom p c= dom (p^<*T2*>) by FINSEQ_1:26;
A20: dom p c= dom (p^<*T1 *>) by FINSEQ_1:26;
A21: (p^<*T2*>).(n+1) = p.(n+1) by A18,FINSEQ_1:def 7;
A22: (p^<*T1*>).(n+1) = p.(n+1) by A18,FINSEQ_1:def 7;
A23: w2.(n+1) = p.(n+1) by A18,A19,A21,FINSEQ_1:def 7;
w1.(n+1) = p.(n+1) by A18,A20,A22,FINSEQ_1:def 7;
then
A24: r in W2 by A3,A4,A5,A6,A12,A13,A14,A23,Def15;
not <*len p*> is_a_prefix_of <*n*>^s by A17,TREES_1:50;
then not <*len p*> is_a_proper_prefix_of <*n*>^s;
hence thesis by A10,A14,A24,TREES_1:def 9;
end;
A25: now
assume
A26: n = len p;
then
A27: (p^<*T1*>).(n+1) = T1 by FINSEQ_1:42;
n < len p+1 by A26,NAT_1:13;
then n+1 in dom (p^<*T1*>) by A3,Lm3;
then
A28: w1.(n+1) = T1 by A27,FINSEQ_1:def 7;
s = s;
hence thesis by A10,A13,A14,A26,A28,TREES_1:def 9;
end;
now
assume that
A29: n >= len p+1 and
A30: n < len p+1+len q;
A31: n+1 >= len p+1+1 by A29,XREAL_1:7;
A32: n+1 <= len p+1+len q by A30,NAT_1:13;
then
A33: w1.(n+1) = q.(n+1-(len p+1)) by A3,A31,FINSEQ_1:23;
w2.(n+1) = q.(n+1-(len p+1)) by A5,A31,A32,FINSEQ_1:23;
then
A34: r in W2 by A3,A4,A5,A6,A12,A13,A14,A33,Def15;
len p <> n by A29,NAT_1:13;
then not <*len p*> is_a_prefix_of <*n*>^s by TREES_1:50;
then not <*len p*> is_a_proper_prefix_of <*n*>^s;
hence thesis by A10,A14,A34,TREES_1:def 9;
end;
hence thesis by A3,A12,A15,A16,A25,FINSEQ_1:22,XXREAL_0:1;
end;
hence thesis by A11,TREES_1:22;
end;
assume r in W;
then
A35: r in W2 & not <*len p*> is_a_proper_prefix_of r or
ex s being FinSequence of NAT st s in T1 & r = <*len p*>^s
by A10,TREES_1:def 9;
assume
A36: not r in W1;
then
A37: r <> {} by Def15;
A38: (p^<*T1*>).(len p+1) = T1 by FINSEQ_1:42;
A39: len p < len p+1 by NAT_1:13;
len p+1 <= len p+1+len q by NAT_1:11;
then
A40: len p < len w2 by A5,A6,NAT_1:13;
len p+1 in dom (p^<*T1*>) by A3,A39,Lm3;
then
A41: w1.(len p+1) = T1 by A38,FINSEQ_1:def 7;
then consider n, s such that
A42: n < len w2 and
A43: s in w2.(n+1) and
A44: r = <*n*>^s by A3,A4,A5,A6,A35,A36,A37,A40,Def15;
reconsider s as FinSequence of NAT by A44,FINSEQ_1:36;
A45: n = len p implies s = {} by A3,A4,A5,A6,A35,A36,A41,A42,A44,Def15,
TREES_1:10;
{} in T1 by TREES_1:22;
then n <> len p by A3,A4,A5,A6,A36,A41,A42,A44,A45,Def15;
then n < len p or n > len p & 1<=1 by XXREAL_0:1;
then 1 <= 1+n & 1+n = n+1 & n+1 <= len p & w1 = p^(<*T1*>^q) &
w2 = p^(<*T2*>^q) or len p+1 < n+1 & n+1 <= len w1
by A3,A4,A5,A6,A42,FINSEQ_1:32,NAT_1:11,13,XREAL_1:6;
then w1.(n+1) = p.(n+1) & w2.(n+1) = p.(n+1) or
w1.(n+1) = q.(n+1-(len p+1)) & w2.(n+1) = q.(n+1-(len p+1))
by A3,A4,A5,A6,Lm1,FINSEQ_1:24;
hence contradiction by A3,A4,A5,A6,A36,A42,A43,A44,Def15;
end;
definition
let n be Nat;
redefine func <*n*> -> FinSequence of NAT;
coherence
proof
reconsider n as Element of NAT by ORDINAL1:def 12;
<*n*> is FinSequence of NAT;
hence thesis;
end;
end;
theorem
for T being Tree holds ^T = elementary_tree 1 with-replacement(<*0*>, T)
proof
let T be Tree;
let p be FinSequence of NAT;
A1: <*T*>.1 = T by FINSEQ_1:40;
A2: len <*T*> = 1 by FINSEQ_1:40;
A3: 0+1 = 1;
A4: {} in T by TREES_1:22;
A5: <*0*> in elementary_tree 1 by TARSKI:def 2,TREES_1:51;
thus p in ^T implies p in elementary_tree 1 with-replacement(<*0*>, T)
proof
assume
A6: p in ^T;
now
assume p <> {};
then consider n,q such that
A7: n < len <*T*> and
A8: q in <*T*>.(n+1) and
A9: p = <*n*>^q by A6,Def15;
reconsider q as FinSequence of NAT by A9,FINSEQ_1:36;
A10: n = 0 by A2,A3,A7,NAT_1:13;
p = <*n*>^q by A9;
hence thesis by A1,A5,A8,A10,TREES_1:def 9;
end;
hence thesis by TREES_1:22;
end;
assume p in elementary_tree 1 with-replacement(<*0*>,T);
then
A11: p in elementary_tree 1 & not <*0*> is_a_proper_prefix_of p or
ex r being FinSequence of NAT st r in T & p = <*0*>^r by A5,TREES_1:def 9;
now
assume p in elementary_tree 1;
then p = {} or p = <*0*> & p^{} = p by FINSEQ_1:34,TARSKI:def 2,TREES_1:51;
hence thesis by A1,A2,A3,A4,Def15;
end;
hence thesis by A1,A2,A3,A11,Def15;
end;
theorem
for T1,T2 being Tree holds
tree(T1,T2) = (elementary_tree 2 with-replacement(<*0*>,T1))
with-replacement (<*1*>, T2)
proof
let T1,T2 be Tree;
let p be FinSequence of NAT;
A1: <*T1,T2*>.1 = T1 by FINSEQ_1:44;
A2: <*T1,T2*>.2 = T2 by FINSEQ_1:44;
A3: len <*T1,T2*> = 2 by FINSEQ_1:44;
A4: 1+1 = 2;
A5: 0+1 = 1;
A6: {} in T1 by TREES_1:22;
A7: {} in T2 by TREES_1:22;
A8: <*0*> in elementary_tree 2 by ENUMSET1:def 1,TREES_1:53;
A9: <*1*> in elementary_tree 2 by ENUMSET1:def 1,TREES_1:53;
not <*0*> is_a_proper_prefix_of <*1*> by TREES_1:52;
then
A10: <*1*> in elementary_tree 2 with-replacement(<*0*>, T1) by A8,A9,
TREES_1:def 9;
thus p in tree(T1,T2) implies
p in (elementary_tree 2 with-replacement (<*0*>,T1))
with-replacement(<*1*>, T2)
proof
assume
A11: p in tree(T1,T2);
now
assume p <> {};
then consider n,q such that
A12: n < len <*T1,T2*> and
A13: q in <*T1,T2*>.(n+1) and
A14: p = <*n*>^q by A11,Def15;
reconsider q as FinSequence of NAT by A14,FINSEQ_1:36;
A15: not <*1*> is_a_prefix_of <*0*>^q by TREES_1:50;
reconsider n as Element of NAT by ORDINAL1:def 12;
A16: now
assume
A17: n = 0;
then
A18: not <*1*> is_a_proper_prefix_of <*n*>^q by A15;
<*n*>^q in elementary_tree 2 with-replacement(<*0*>, T1) by A1,A8,A13
,A17,TREES_1:def 9;
hence thesis by A10,A14,A18,TREES_1:def 9;
end;
n <= 0+1 by A3,A4,A12,NAT_1:13;
then n = 1 & (n = 1 implies
<*n*>^q in (elementary_tree 2 with-replacement (<*0*>,T1))
with-replacement(<*1*>,T2)) or
n >= 0 & n <= 0 by A2,A10,A13,NAT_1:8,TREES_1:def 9;
hence thesis by A14,A16;
end;
hence thesis by TREES_1:22;
end;
assume p in (elementary_tree 2 with-replacement (<*0*>,T1))
with-replacement(<*1*>,T2);
then
A19: p in elementary_tree 2 with-replacement(<*0*>,T1) &
not <*1*> is_a_proper_prefix_of p or ex r being FinSequence of NAT st r in
T2 & p = <*1*>^r by A10,TREES_1:def 9;
now
assume p in elementary_tree 2 with-replacement(<*0*>,T1);
then
A20: p in elementary_tree 2 & not <*0*> is_a_proper_prefix_of p or
ex r being FinSequence of NAT st r in T1 & p = <*0*>^r
by A8,TREES_1:def 9;
now
assume p in elementary_tree 2;
then
A21: p = {} or p = <*0*> or p = <*1*> by ENUMSET1:def 1,TREES_1:53;
p^{} = p by FINSEQ_1:34;
hence thesis by A1,A2,A3,A4,A5,A6,A7,A21,Def15;
end;
hence thesis by A1,A3,A5,A20,Def15;
end;
hence thesis by A2,A3,A4,A19,Def15;
end;
registration
let p be FinTree-yielding FinSequence;
cluster tree(p) -> finite;
coherence
proof
defpred X[set] means
for p being FinTree-yielding FinSequence st len p = $1 holds
tree(p) is finite Tree;
A1: X[0]
proof
let p be FinTree-yielding FinSequence;
assume len p = 0;
then p = {};
hence thesis by Th52;
end;
A2: X[n] implies X[n+1]
proof
assume
A3: for p being FinTree-yielding FinSequence st len p = n holds
tree(p) is finite Tree;
let p be FinTree-yielding FinSequence such that
A4: len p = n+1;
p <> {} by A4;
then consider q being FinSequence, x being object such that
A5: p = q^<*x*> by FINSEQ_1:46;
reconsider q as FinTree-yielding FinSequence by A5,Th26;
len <*x*> = 1 by FINSEQ_1:40;
then
A6: len p = len q+1 by A5,FINSEQ_1:22;
then tree(q) is finite by A3,A4;
then reconsider W = tree(q) \/ elementary_tree (n+1) as finite Tree;
<*x*> is FinTree-yielding by A5,Th26;
then reconsider x as finite Tree by Th29;
n < n+1 by NAT_1:13;
then <*n*> in elementary_tree (n+1) by TREES_1:28;
then reconsider r = <*n*> as Element of W by XBOOLE_0:def 3;
tree(p) = W with-replacement(r, x) by A4,A5,A6,Th55;
hence thesis;
end;
X[n] from NAT_1:sch 2(A1,A2);
then len p = len p implies thesis;
hence thesis;
end;
end;
registration
let T be finite Tree;
cluster ^T -> finite;
coherence;
end;
registration
let T1,T2 be finite Tree;
cluster tree(T1,T2) -> finite;
coherence;
end;
theorem Th60:
for T being Tree, x being object holds x in ^T iff x = {} or
ex p st p in T & x = <*0*>^p
proof
let T be Tree;
let x;
set p = <*T*>;
A1: len p = 1 by FINSEQ_1:40;
A2: p.1 = T by FINSEQ_1:40;
thus x in ^T & x <> {} implies ex p st p in T & x = <*0*>^p
proof
assume that
A3: x in ^T and
A4: x <> {};
consider n, q such that
A5: n < len p and
A6: q in p.(n+1) and
A7: x = <*n*>^q by A3,A4,Def15;
0+1 = 1;
then n = 0 by A1,A5,NAT_1:13;
hence thesis by A2,A6,A7;
end;
0 < 0+1;
hence thesis by A1,A2,Def15;
end;
theorem Th61:
for T being Tree, p being FinSequence holds p in T iff <*0*>^p in ^T
proof
let T be Tree, p be FinSequence;
thus p in T implies <*0*>^p in ^T by Th60;
assume <*0*>^p in ^T;
then consider n,q such that
n < len <*T*> and
A1: q in <*T*>.(n+1) and
A2: <*0*>^p = <*n*>^q by Def15;
A3: (<*0*>^p).1 = 0 by FINSEQ_1:41;
A4: (<*n*>^q).1 = n by FINSEQ_1:41;
then p = q by A2,A3,FINSEQ_1:33;
hence thesis by A1,A2,A3,A4,FINSEQ_1:40;
end;
theorem
for T being Tree holds elementary_tree 1 c= ^T
proof
let T be Tree;
let x be object;
assume x in elementary_tree 1;
then reconsider p = x as Element of elementary_tree 1;
p = {} or p = <*0*> & {} in T & <*0*>^{} = <*0*>
by FINSEQ_1:34,TARSKI:def 2,TREES_1:22,51;
hence thesis by Th60;
end;
theorem
for T1,T2 being Tree st T1 c= T2 holds ^T1 c= ^T2
proof
let T1,T2 be Tree such that
A1: T1 c= T2;
let x be object;
assume x in ^T1;
then reconsider p = x as Element of ^T1;
p = {} or ex q st q in T1 & p = <*0*>^q by Th60;
hence thesis by A1,Th60;
end;
theorem
for T1,T2 being Tree st ^T1 = ^T2 holds T1 = T2
proof
let T1,T2 be Tree such that
A1: ^T1 = ^T2;
let p be FinSequence of NAT;
p in T1 iff <*0*>^p in ^T1 by Th61;
hence thesis by A1,Th61;
end;
theorem
for T being Tree holds (^T)|<*0*> = T
proof
let T be Tree;
set p = <*T*>;
A1: len p = 1 by FINSEQ_1:40;
A2: p.1 = T by FINSEQ_1:40;
0+1 = 1;
hence thesis by A1,A2,Th49;
end;
theorem
for T1,T2 being Tree holds ^T1 with-replacement (<*0*>,T2) = ^T2
proof
let T1,T2 be Tree;
A1: len {} = 0;
A2: <*T1*> = {}^<*T1*> by FINSEQ_1:34;
A3: <*T2*> = {}^<*T2*> by FINSEQ_1:34;
A4: <*T1*> = <*T1*>^{} by FINSEQ_1:34;
<*T2*> = <*T2*>^{} by FINSEQ_1:34;
hence thesis by A1,A2,A3,A4,Th57;
end;
theorem
^elementary_tree 0 = elementary_tree 1
proof
set T = elementary_tree 0;
thus ^T = tree(1 |-> T) by FINSEQ_2:59
.= elementary_tree 1 by Th54;
end;
theorem Th68:
for T1,T2 being Tree, x being object holds x in tree(T1,T2) iff x = {} or
ex p st p in T1 & x = <*0*>^p or p in T2 & x = <*1*>^p
proof
let T1, T2 be Tree;
let x;
set p = <*T1,T2*>;
A1: len p = 2 by FINSEQ_1:44;
A2: p.1 = T1 by FINSEQ_1:44;
A3: p.2 = T2 by FINSEQ_1:44;
thus x in tree(T1,T2) & x <> {} implies
ex p st p in T1 & x = <*0*>^p or p in T2 & x = <*1*>^p
proof
assume that
A4: x in tree(T1,T2) and
A5: x <> {};
consider n, q such that
A6: n < len p and
A7: q in p.(n+1) and
A8: x = <*n*>^q by A4,A5,Def15;
1+1 = 2;
then n <= 1 by A1,A6,NAT_1:13;
then n = 1 or n < 0+1 by XXREAL_0:1;
then n = 1 or n = 0 by NAT_1:13;
hence thesis by A2,A3,A7,A8;
end;
now
given q such that
A9: q in T1 & x = <*0*>^q or q in T2 & x = <*1*>^q;
x = <*0*>^q or x <> <*0*>^q;
then consider n such that
A10: n = 0 & x = <*0*>^q or n = 1 & x <> <*0*>^q;
take n,q;
thus n < len p by A1,A10;
(<*0*>^q).1 = 0 by FINSEQ_1:41;
hence q in p.(n+1) & x = <*n*>^q by A9,A10,FINSEQ_1:41,44;
end;
hence thesis by Def15;
end;
theorem Th69:
for T1,T2 being Tree, p being FinSequence holds
p in T1 iff <*0*>^p in tree(T1,T2)
proof
let T1,T2 be Tree;
A1: <*T1,T2*> = <*T1*>^<*T2*> by FINSEQ_1:def 9;
A2: <*T1*> = {}^<*T1*> by FINSEQ_1:34;
len {} = 0;
hence thesis by A1,A2,Th51;
end;
theorem Th70:
for T1,T2 being Tree, p being FinSequence holds
p in T2 iff <*1*>^p in tree(T1,T2)
proof
let T1,T2 be Tree;
A1: <*T1,T2*> = <*T1*>^<*T2*> by FINSEQ_1:def 9;
A2: len <*T1*> = 1 by FINSEQ_1:40;
<*T1,T2*> = <*T1,T2*>^{} by FINSEQ_1:34;
hence thesis by A1,A2,Th51;
end;
theorem
for T1,T2 being Tree holds elementary_tree 2 c= tree(T1,T2)
proof
let T1,T2 be Tree;
let x be object;
assume x in elementary_tree 2;
then reconsider p = x as Element of elementary_tree 2;
p = {} or p = <*0*> & {} in T1 & <*0*>^{} = <*0*> or
p = <*1*> & {} in T2 & <*1*>^{} = <*1*>
by ENUMSET1:def 1,FINSEQ_1:34,TREES_1:22,53;
hence thesis by Th68;
end;
theorem
for T1,T2, W1,W2 being Tree st T1 c= W1 & T2 c= W2 holds
tree(T1,T2) c= tree(W1,W2)
proof
let T1,T2, W1,W2 be Tree such that
A1: T1 c= W1 and
A2: T2 c= W2;
let x be object;
assume x in tree(T1,T2);
then reconsider p = x as Element of tree(T1,T2);
p = {} or ex q st q in T1 & p = <*0*>^q or q in T2 & p = <*1*>^q by Th68;
hence thesis by A1,A2,Th68;
end;
theorem
for T1,T2, W1,W2 being Tree st tree(T1,T2) = tree(W1,W2) holds
T1 = W1 & T2 = W2
proof
let T1,T2, W1,W2 be Tree such that
A1: tree(T1,T2) = tree(W1,W2);
now
let p;
p in T1 iff <*0*>^p in tree(T1,T2) by Th69;
hence p in T1 iff p in W1 by A1,Th69;
end;
hence for p being FinSequence of NAT holds p in T1 iff p in W1;
let p be FinSequence of NAT;
p in T2 iff <*1*>^p in tree(T1,T2) by Th70;
hence thesis by A1,Th70;
end;
theorem
for T1,T2 being Tree holds tree(T1,T2)|<*0*> = T1 & tree(T1,T2)|<*1*> = T2
proof
let T1, T2 be Tree;
set p = <*T1,T2*>;
A1: len p = 2 by FINSEQ_1:44;
A2: p.1 = T1 by FINSEQ_1:44;
A3: p.2 = T2 by FINSEQ_1:44;
A4: 0+1 = 1;
1+1 = 2;
hence thesis by A1,A2,A3,A4,Th49;
end;
theorem
for T,T1,T2 being Tree holds
tree(T1,T2) with-replacement (<*0*>, T) = tree(T,T2) &
tree(T1,T2) with-replacement (<*1*>, T) = tree(T1,T)
proof
let T,T1,T2 be Tree;
A1: len {} = 0;
A2: <*T1*> = {}^<*T1*> by FINSEQ_1:34;
A3: <*T*> = {}^<*T*> by FINSEQ_1:34;
A4: <*T1,T2*>^{} = <*T1,T2*> by FINSEQ_1:34;
A5: <*T1,T*>^{} = <*T1,T*> by FINSEQ_1:34;
A6: len <*T1*> = 1 by FINSEQ_1:40;
A7: <*T1,T2*> = <*T1*>^<*T2*> by FINSEQ_1:def 9;
A8: <*T1,T*> = <*T1*>^<*T*> by FINSEQ_1:def 9;
<*T,T2*> = <*T*>^<*T2*> by FINSEQ_1:def 9;
hence thesis by A1,A2,A3,A4,A5,A6,A7,A8,Th57;
end;
theorem
tree(elementary_tree 0, elementary_tree 0) = elementary_tree 2
proof
set T = elementary_tree 0;
thus tree(T,T) = tree(2 |-> T) by FINSEQ_2:61
.= elementary_tree 2 by Th54;
end;
reserve w for FinTree-yielding FinSequence;
theorem Th77:
for w st for t being finite Tree st t in rng w holds height t <= n holds
height tree(w) <= n+1
proof
let w such that
A1: for t being finite Tree st t in rng w holds height t <= n;
consider p being FinSequence of NAT such that
A2: p in tree(w) and
A3: len p = height tree(w) by TREES_1:def 12;
A4: p = {} & len {} = 0 or ex n,q st n < len w & q in w.(n+1) & p = <*n*>^q
by A2,Def15;
now
given k,q such that
A5: k < len w and
A6: q in w.(k+1) and
A7: p = <*k*>^q;
A8: w.(k+1) in rng w by A5,Lm3;
rng w is constituted-FinTrees by Def10;
then reconsider t = w.(k+1) as finite Tree by A8;
reconsider q as FinSequence of NAT by A7,FINSEQ_1:36;
A9: len q <= height t by A6,TREES_1:def 12;
A10: height t <= n by A1,A5,Lm3;
A11: len <*k*> = 1 by FINSEQ_1:40;
A12: len q <= n by A9,A10,XXREAL_0:2;
len p = 1+len q by A7,A11,FINSEQ_1:22;
hence thesis by A3,A12,XREAL_1:7;
end;
hence thesis by A3,A4;
end;
theorem Th78:
for t being finite Tree st t in rng w holds height tree(w) > height t
proof
let t be finite Tree;
assume t in rng w;
then consider x being object such that
A1: x in dom w and
A2: t = w.x by FUNCT_1:def 3;
reconsider x as Element of NAT by A1;
A3: 1 <= x by A1,FINSEQ_3:25;
A4: len w >= x by A1,FINSEQ_3:25;
consider n be Nat such that
A5: x = 1+n by A3,NAT_1:10;
reconsider n as Element of NAT by ORDINAL1:def 12;
A6: n < len w by A4,A5,NAT_1:13;
A7: {} in t by TREES_1:22;
A8: <*n*>^{} = <*n*> by FINSEQ_1:34;
A9: t = (tree(w))|<*n*> by A2,A5,A6,Th49;
<*n*> in tree(w) by A2,A5,A6,A7,A8,Def15;
hence thesis by A9,TREES_1:48;
end;
theorem Th79:
for t being finite Tree st t in rng w &
for t9 being finite Tree st t9 in rng w holds height t9 <= height t holds
height tree(w) = (height t) + 1
proof
let t be finite Tree such that
A1: t in rng w and
A2: for t9 being finite Tree st t9 in rng w holds height t9 <= height t;
A3: height tree(w) > height t by A1,Th78;
A4: height tree(w) <= (height t) + 1 by A2,Th77;
height tree(w) >= (height t) + 1 by A3,NAT_1:13;
hence thesis by A4,XXREAL_0:1;
end;
theorem
for T being finite Tree holds height ^T = (height T) + 1
proof
let T be finite Tree;
set m = height T;
A1: rng <*T*> = {T} by FINSEQ_1:38;
A2: T in {T} by TARSKI:def 1;
for t being finite Tree st t in rng <*T*> holds height t <= m by A1,
TARSKI:def 1;
hence thesis by A1,A2,Th79;
end;
theorem
for T1,T2 being finite Tree holds
height tree(T1,T2) = max(height T1, height T2)+1
proof
let T1,T2 be finite Tree;
set m = max(height T1, height T2);
A1: rng <*T1,T2*> = {T1,T2} by FINSEQ_2:127;
A2: T1 in {T1, T2} by TARSKI:def 2;
A3: T2 in {T1,T2} by TARSKI:def 2;
A4: m = height T1 or m = height T2 by XXREAL_0:def 10;
now
let t be finite Tree;
assume t in rng <*T1,T2*>;
then t = T1 or t = T2 by A1,TARSKI:def 2;
hence height t <= m by XXREAL_0:25;
end;
hence thesis by A1,A2,A3,A4,Th79;
end;
begin :: Addenda
:: from DTCONSTR
registration
let D be non empty set, t be Element of FinTrees D;
cluster dom t -> finite;
coherence by Def8;
end;
definition
let p be FinSequence such that
A1: p is DTree-yielding;
defpred P[Nat, object] means
ex T being DecoratedTree st T = p.$1 & $2 = T.{};
func roots p -> FinSequence means
dom it = dom p & for i being Element of NAT st i in dom p
ex T being DecoratedTree st T = p.i & it.i = T.{};
existence
proof
A2: Seg len p = dom p by FINSEQ_1:def 3;
A3: for k being Nat st k in Seg len p ex x being object st P[k,x]
proof
let k be Nat;
assume k in Seg len p;
then
A4: p.k in rng p by A2,FUNCT_1:def 3;
rng p is constituted-DTrees by A1;
then reconsider T = p.k as DecoratedTree by A4;
take T.{}, T;
thus thesis;
end;
consider q being FinSequence such that
A5: dom q = Seg len p & for k being Nat st k in Seg len p
holds P[k,q.k] from FINSEQ_1:sch 1(A3);
take q;
thus dom q = dom p by A5,FINSEQ_1:def 3;
thus thesis by A2,A5;
end;
uniqueness
proof
let x1, x2 be FinSequence such that
A6: dom x1 = dom p and
A7: for i being Element of NAT st i in dom p holds P[i,x1 .i] and
A8: dom x2 = dom p and
A9: for i being Element of NAT st i in dom p holds P[ i,x2.i];
now
let k be Nat;
assume
A10: k in dom p;
then
A11: P[k,x1.k] by A7;
P[k,x2.k] by A9,A10;
hence x1.k = x2.k by A11;
end;
hence thesis by A6,A8;
end;
end;