let A be set ; :: thesis: for X being non empty set
for f being Function of X,[:(Fin A),(Fin A):]
for B being Element of Fin X
for c being Element of [:(Fin A),(Fin A):] st ( for x being Element of X st x in B holds
f . x c= c ) holds
FinPairUnion B,f c= c

let X be non empty set ; :: thesis: for f being Function of X,[:(Fin A),(Fin A):]
for B being Element of Fin X
for c being Element of [:(Fin A),(Fin A):] st ( for x being Element of X st x in B holds
f . x c= c ) holds
FinPairUnion B,f c= c

let f be Function of X,[:(Fin A),(Fin A):]; :: thesis: for B being Element of Fin X
for c being Element of [:(Fin A),(Fin A):] st ( for x being Element of X st x in B holds
f . x c= c ) holds
FinPairUnion B,f c= c

let B be Element of Fin X; :: thesis: for c being Element of [:(Fin A),(Fin A):] st ( for x being Element of X st x in B holds
f . x c= c ) holds
FinPairUnion B,f c= c

let c be Element of [:(Fin A),(Fin A):]; :: thesis: ( ( for x being Element of X st x in B holds
f . x c= c ) implies FinPairUnion B,f c= c )

defpred S1[ Element of Fin X] means ( $1 c= B implies (FinPairUnion A) $$ $1,f c= c );
assume A1: for x being Element of X st x in B holds
f . x c= c ; :: thesis: FinPairUnion B,f c= c
A2: now
let C be Element of Fin X; :: thesis: for b being Element of X st S1[C] holds
S1[C \/ {.b.}]

let b be Element of X; :: thesis: ( S1[C] implies S1[C \/ {.b.}] )
assume A3: S1[C] ; :: thesis: S1[C \/ {.b.}]
now
assume A4: C \/ {b} c= B ; :: thesis: (FinPairUnion A) $$ (C \/ {.b.}),f c= c
then {b} c= B by XBOOLE_1:11;
then b in B by ZFMISC_1:37;
then A5: f . b c= c by A1;
(FinPairUnion A) $$ (C \/ {.b.}),f = (FinPairUnion A) . ((FinPairUnion A) $$ C,f),(f . b) by Th37, SETWISEO:41
.= ((FinPairUnion A) $$ C,f) \/ (f . b) by Def6 ;
hence (FinPairUnion A) $$ (C \/ {.b.}),f c= c by A3, A4, A5, Th25, XBOOLE_1:11; :: thesis: verum
end;
hence S1[C \/ {.b.}] ; :: thesis: verum
end;
A6: S1[ {}. X]
proof end;
for C being Element of Fin X holds S1[C] from SETWISEO:sch 4(A6, A2);
hence FinPairUnion B,f c= c ; :: thesis: verum