reconsider n1 = n as Element of NAT by ORDINAL1:def 12;

reconsider T = Table2 (m,e,f,n) as Element of NAT ;

defpred S_{1}[ Nat, set , set ] means ex i1, i2 being Nat st

( i1 = $2 & i2 = $3 & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n1 -' $1)))) mod f );

A2: for i being Nat st 1 <= i & i < n1 holds

for x being Element of NAT ex y being Element of NAT st S_{1}[i,x,y]

A4: ( len r = n1 & ( r . 1 = T or n1 = 0 ) ) and

A5: for i being Nat st 1 <= i & i < n1 holds

S_{1}[i,r . i,r . (i + 1)]
from RECDEF_1:sch 4(A2);

reconsider r = r as Tuple of n, NAT by A4, CARD_1:def 7;

take r ; :: thesis: ( r . 1 = Table2 (m,e,f,n) & ( for i being Nat st 1 <= i & i <= n - 1 holds

ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f ) ) )

thus r . 1 = Table2 (m,e,f,n) by A1, A4; :: thesis: for i being Nat st 1 <= i & i <= n - 1 holds

ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f )

let i be Nat; :: thesis: ( 1 <= i & i <= n - 1 implies ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f ) )

assume A6: ( 1 <= i & i <= n - 1 ) ; :: thesis: ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f )

thus ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f ) by A5, A6, XREAL_1:147; :: thesis: verum

reconsider T = Table2 (m,e,f,n) as Element of NAT ;

defpred S

( i1 = $2 & i2 = $3 & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n1 -' $1)))) mod f );

A2: for i being Nat st 1 <= i & i < n1 holds

for x being Element of NAT ex y being Element of NAT st S

proof

consider r being FinSequence of NAT such that
let i be Nat; :: thesis: ( 1 <= i & i < n1 implies for x being Element of NAT ex y being Element of NAT st S_{1}[i,x,y] )

assume that

1 <= i and

i < n1 ; :: thesis: for x being Element of NAT ex y being Element of NAT st S_{1}[i,x,y]

let x be Element of NAT ; :: thesis: ex y being Element of NAT st S_{1}[i,x,y]

reconsider x = x as Element of NAT ;

consider y being Element of NAT such that

A3: y = (((x |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f ;

reconsider z = y as Element of NAT ;

take z ; :: thesis: S_{1}[i,x,z]

thus S_{1}[i,x,z]
by A3; :: thesis: verum

end;assume that

1 <= i and

i < n1 ; :: thesis: for x being Element of NAT ex y being Element of NAT st S

let x be Element of NAT ; :: thesis: ex y being Element of NAT st S

reconsider x = x as Element of NAT ;

consider y being Element of NAT such that

A3: y = (((x |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f ;

reconsider z = y as Element of NAT ;

take z ; :: thesis: S

thus S

A4: ( len r = n1 & ( r . 1 = T or n1 = 0 ) ) and

A5: for i being Nat st 1 <= i & i < n1 holds

S

reconsider r = r as Tuple of n, NAT by A4, CARD_1:def 7;

take r ; :: thesis: ( r . 1 = Table2 (m,e,f,n) & ( for i being Nat st 1 <= i & i <= n - 1 holds

ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f ) ) )

thus r . 1 = Table2 (m,e,f,n) by A1, A4; :: thesis: for i being Nat st 1 <= i & i <= n - 1 holds

ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f )

let i be Nat; :: thesis: ( 1 <= i & i <= n - 1 implies ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f ) )

assume A6: ( 1 <= i & i <= n - 1 ) ; :: thesis: ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f )

thus ex i1, i2 being Nat st

( i1 = r . i & i2 = r . (i + 1) & i2 = (((i1 |^ (Radix k)) mod f) * (Table2 (m,e,f,(n -' i)))) mod f ) by A5, A6, XREAL_1:147; :: thesis: verum