reserve f,g,h for Function,
  A for set;
reserve F for Function,
  B,x,y,y1,y2,z for set;
reserve x,z for object;

theorem Th24:
  f|A = g|A implies (F.:(h,f))|A = (F.:(h,g))|A
proof
  assume
A1: f|A = g|A;
  thus (F.:(h,f))|A = F*<:h,f:>|A by RELAT_1:83
    .= F*<:h,f|A:> by Th6
    .= F*<:h,g:>|A by A1,Th6
    .= (F.:(h,g))|A by RELAT_1:83;
end;
