let f1, f2 be Function of NAT,S; :: thesis: ( ( for n being Element of NAT holds f1 . n = PathChange (pai1,pai2,k,n) ) & ( for n being Element of NAT holds f2 . n = PathChange (pai1,pai2,k,n) ) implies f1 = f2 )
assume that
A5: for n being Element of NAT holds f1 . n = PathChange (pai1,pai2,k,n) and
A6: for n being Element of NAT holds f2 . n = PathChange (pai1,pai2,k,n) ; :: thesis: f1 = f2
for x being set st x in NAT holds
f1 . x = f2 . x

proof

let x be set ; :: thesis: ( x in NAT implies f1 . x = f2 . x )
assume x in NAT ; :: thesis: f1 . x = f2 . x
then reconsider x = x as Element of NAT ;
f1 . x = PathChange (pai1,pai2,k,x) by A5
.= f2 . x by A6 ;
hence f1 . x = f2 . x ; :: thesis: verum

end;

hence f1 = f2 by FUNCT_2:12; :: thesis: verum