let f be FinSequence of (); :: thesis: for Q being Subset of ()
for q being Point of ()
for i, j being Nat st L~ f meets Q & f is being_S-Seq & Q is closed & Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) in LSeg (f,i) & 1 <= i & i + 1 <= len f & q in LSeg (f,j) & 1 <= j & j + 1 <= len f & q in Q & Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) <> q holds
( i >= j & ( i = j implies LE q, Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q),f /. i,f /. (i + 1) ) )

let Q be Subset of (); :: thesis: for q being Point of ()
for i, j being Nat st L~ f meets Q & f is being_S-Seq & Q is closed & Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) in LSeg (f,i) & 1 <= i & i + 1 <= len f & q in LSeg (f,j) & 1 <= j & j + 1 <= len f & q in Q & Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) <> q holds
( i >= j & ( i = j implies LE q, Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q),f /. i,f /. (i + 1) ) )

let q be Point of (); :: thesis: for i, j being Nat st L~ f meets Q & f is being_S-Seq & Q is closed & Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) in LSeg (f,i) & 1 <= i & i + 1 <= len f & q in LSeg (f,j) & 1 <= j & j + 1 <= len f & q in Q & Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) <> q holds
( i >= j & ( i = j implies LE q, Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q),f /. i,f /. (i + 1) ) )

let i, j be Nat; :: thesis: ( L~ f meets Q & f is being_S-Seq & Q is closed & Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) in LSeg (f,i) & 1 <= i & i + 1 <= len f & q in LSeg (f,j) & 1 <= j & j + 1 <= len f & q in Q & Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) <> q implies ( i >= j & ( i = j implies LE q, Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q),f /. i,f /. (i + 1) ) ) )
assume that
A1: L~ f meets Q and
A2: f is being_S-Seq and
A3: Q is closed and
A4: Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) in LSeg (f,i) and
A5: ( 1 <= i & i + 1 <= len f ) and
A6: q in LSeg (f,j) and
A7: 1 <= j and
A8: j + 1 <= len f and
A9: q in Q and
A10: Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q) <> q ; :: thesis: ( i >= j & ( i = j implies LE q, Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q),f /. i,f /. (i + 1) ) )
reconsider P = L~ f as non empty Subset of () by ;
set q1 = Last_Point (P,(f /. 1),(f /. (len f)),Q);
set p2 = f /. (i + 1);
A11: q in L~ f by ;
thus i >= j :: thesis: ( i = j implies LE q, Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q),f /. i,f /. (i + 1) )
proof
assume j > i ; :: thesis: contradiction
then A12: i + 1 <= j by NAT_1:13;
j <= j + 1 by NAT_1:11;
then ( 1 <= i + 1 & j <= len f ) by ;
then A13: LE f /. (i + 1),f /. j,P,f /. 1,f /. (len f) by ;
LE f /. j,q,P,f /. 1,f /. (len f) by A2, A6, A7, A8, Th25;
then A14: LE f /. (i + 1),q,P,f /. 1,f /. (len f) by ;
(L~ f) /\ Q is closed by ;
then A15: LE q, Last_Point (P,(f /. 1),(f /. (len f)),Q),P,f /. 1,f /. (len f) by A2, A9, A11, Th16;
LE Last_Point (P,(f /. 1),(f /. (len f)),Q),f /. (i + 1),P,f /. 1,f /. (len f) by A2, A4, A5, Th26;
then LE Last_Point (P,(f /. 1),(f /. (len f)),Q),q,P,f /. 1,f /. (len f) by ;
hence contradiction by A2, A10, A15, Th12, TOPREAL1:25; :: thesis: verum
end;
assume i = j ; :: thesis: LE q, Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q),f /. i,f /. (i + 1)
hence LE q, Last_Point ((L~ f),(f /. 1),(f /. (len f)),Q),f /. i,f /. (i + 1) by A1, A2, A3, A4, A5, A6, A9, Lm4; :: thesis: verum