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
  f|A = g|A implies (F[;](x,f))|A = (F[;](x,g))|A
proof
  assume
A1: f|A = g|A;
  dom f /\ A = dom (f|A) by RELAT_1:61
    .= dom g /\ A by A1,RELAT_1:61;
  then
A2: (dom f --> x)|A = dom g /\ A --> x by Th12
    .= (dom g -->x)|A by Th12;
  thus (F[;](x,f))|A = (F.:(dom f --> x, f))|A
    .= (F.:(dom f --> x, g))|A by A1,Th24
    .= (F.:(dom g --> x, g))|A by A2,Th23
    .= (F[;](x,g))|A;
end;
