now for IT1, IT2 being Function of E,E st ex PF1 being FinSequence of Funcs (E,E) st
( IT1 = PF1 . (len F) & len PF1 = len F & PF1 . 1 = A . (F . 1) & ( for k being Nat st k <> 0 & k < len F holds
ex f, g being Function of E,E st
( f = PF1 . k & g = A . (F . (k + 1)) & PF1 . (k + 1) = f * g ) ) ) & ex PF2 being FinSequence of Funcs (E,E) st
( IT2 = PF2 . (len F) & len PF2 = len F & PF2 . 1 = A . (F . 1) & ( for k being Nat st k <> 0 & k < len F holds
ex f, g being Function of E,E st
( f = PF2 . k & g = A . (F . (k + 1)) & PF2 . (k + 1) = f * g ) ) ) holds
IT1 = IT2let IT1,
IT2 be
Function of
E,
E;
( ex PF1 being FinSequence of Funcs (E,E) st
( IT1 = PF1 . (len F) & len PF1 = len F & PF1 . 1 = A . (F . 1) & ( for k being Nat st k <> 0 & k < len F holds
ex f, g being Function of E,E st
( f = PF1 . k & g = A . (F . (k + 1)) & PF1 . (k + 1) = f * g ) ) ) & ex PF2 being FinSequence of Funcs (E,E) st
( IT2 = PF2 . (len F) & len PF2 = len F & PF2 . 1 = A . (F . 1) & ( for k being Nat st k <> 0 & k < len F holds
ex f, g being Function of E,E st
( f = PF2 . k & g = A . (F . (k + 1)) & PF2 . (k + 1) = f * g ) ) ) implies IT1 = IT2 )given PF1 being
FinSequence of
Funcs (
E,
E)
such that A32:
IT1 = PF1 . (len F)
and A33:
len PF1 = len F
and A34:
PF1 . 1
= A . (F . 1)
and A35:
for
k being
Nat st
k <> 0 &
k < len F holds
ex
f,
g being
Function of
E,
E st
(
f = PF1 . k &
g = A . (F . (k + 1)) &
PF1 . (k + 1) = f * g )
;
( ex PF2 being FinSequence of Funcs (E,E) st
( IT2 = PF2 . (len F) & len PF2 = len F & PF2 . 1 = A . (F . 1) & ( for k being Nat st k <> 0 & k < len F holds
ex f, g being Function of E,E st
( f = PF2 . k & g = A . (F . (k + 1)) & PF2 . (k + 1) = f * g ) ) ) implies IT1 = IT2 )given PF2 being
FinSequence of
Funcs (
E,
E)
such that A36:
(
IT2 = PF2 . (len F) &
len PF2 = len F )
and A37:
PF2 . 1
= A . (F . 1)
and A38:
for
k being
Nat st
k <> 0 &
k < len F holds
ex
f,
g being
Function of
E,
E st
(
f = PF2 . k &
g = A . (F . (k + 1)) &
PF2 . (k + 1) = f * g )
;
IT1 = IT2defpred S1[
Nat]
means ( 1
<= $1 & $1
<= len PF1 implies
PF1 . $1
= PF2 . $1 );
A39:
for
k being
Nat st
S1[
k] holds
S1[
k + 1]
proof
let k be
Nat;
( S1[k] implies S1[k + 1] )
assume A40:
S1[
k]
;
S1[k + 1]
now ( 1 <= k + 1 & k + 1 <= len PF1 implies PF1 . (k + 1) = PF2 . (k + 1) )assume
1
<= k + 1
;
( k + 1 <= len PF1 implies PF1 . (k + 1) = PF2 . (k + 1) )assume A41:
k + 1
<= len PF1
;
PF1 . (k + 1) = PF2 . (k + 1)then A42:
k < len PF1
by NAT_1:13;
per cases
( k = 0 or k <> 0 )
;
suppose A43:
k <> 0
;
PF1 . (k + 1) = PF2 . (k + 1)then A44:
0 + 1
< k + 1
by XREAL_1:6;
( ex
f1,
g1 being
Function of
E,
E st
(
f1 = PF1 . k &
g1 = A . (F . (k + 1)) &
PF1 . (k + 1) = f1 * g1 ) & ex
f2,
g2 being
Function of
E,
E st
(
f2 = PF2 . k &
g2 = A . (F . (k + 1)) &
PF2 . (k + 1) = f2 * g2 ) )
by A33, A35, A38, A42, A43;
hence
PF1 . (k + 1) = PF2 . (k + 1)
by A40, A41, A44, NAT_1:13;
verum end; end; end;
hence
S1[
k + 1]
;
verum
end; A45:
S1[
0 ]
;
for
k being
Nat holds
S1[
k]
from NAT_1:sch 2(A45, A39);
hence
IT1 = IT2
by A32, A33, A36, FINSEQ_1:14;
verum end;
hence
for b1, b2 being Function of E,E holds
( ( len F = 0 & b1 = id E & b2 = id E implies b1 = b2 ) & ( not len F = 0 & ex PF being FinSequence of Funcs (E,E) st
( b1 = PF . (len F) & len PF = len F & PF . 1 = A . (F . 1) & ( for n being Nat st n <> 0 & n < len F holds
ex f, g being Function of E,E st
( f = PF . n & g = A . (F . (n + 1)) & PF . (n + 1) = f * g ) ) ) & ex PF being FinSequence of Funcs (E,E) st
( b2 = PF . (len F) & len PF = len F & PF . 1 = A . (F . 1) & ( for n being Nat st n <> 0 & n < len F holds
ex f, g being Function of E,E st
( f = PF . n & g = A . (F . (n + 1)) & PF . (n + 1) = f * g ) ) ) implies b1 = b2 ) )
; verum