defpred S1[ Nat, set ] means $2 = FreeOpNSG ($1,f,D);
set A = TS (DTConUA (f,D));
A1:
for n being Nat st n in Seg (len f) holds
ex x being Element of PFuncs (((TS (DTConUA (f,D))) *),(TS (DTConUA (f,D)))) st S1[n,x]
proof
let n be
Nat;
( n in Seg (len f) implies ex x being Element of PFuncs (((TS (DTConUA (f,D))) *),(TS (DTConUA (f,D)))) st S1[n,x] )
assume
n in Seg (len f)
;
ex x being Element of PFuncs (((TS (DTConUA (f,D))) *),(TS (DTConUA (f,D)))) st S1[n,x]
reconsider O =
FreeOpNSG (
n,
f,
D) as
Element of
PFuncs (
((TS (DTConUA (f,D))) *),
(TS (DTConUA (f,D))))
by PARTFUN1:45;
take
O
;
S1[n,O]
thus
S1[
n,
O]
;
verum
end;
consider p being FinSequence of PFuncs (((TS (DTConUA (f,D))) *),(TS (DTConUA (f,D)))) such that
A2:
( dom p = Seg (len f) & ( for n being Nat st n in Seg (len f) holds
S1[n,p . n] ) )
from FINSEQ_1:sch 5(A1);
reconsider p = p as PFuncFinSequence of (TS (DTConUA (f,D))) ;
take
p
; ( len p = len f & ( for n being Nat st n in dom p holds
p . n = FreeOpNSG (n,f,D) ) )
thus
len p = len f
by A2, FINSEQ_1:def 3; for n being Nat st n in dom p holds
p . n = FreeOpNSG (n,f,D)
let n be Nat; ( n in dom p implies p . n = FreeOpNSG (n,f,D) )
assume
n in dom p
; p . n = FreeOpNSG (n,f,D)
hence
p . n = FreeOpNSG (n,f,D)
by A2; verum