theorem
  z in A implies (A --> x)*(y .--> z) = y .--> x
 proof assume
A1: z in A;
A2:  dom(y .--> z) = {y}
       .= dom(y .--> x);
   rng(y .--> z) = {z} by Th88;
   then rng(y .--> z) c= dom(A --> x) by A1,ZFMISC_1:31;
  hence dom((A --> x)*(y .--> z)) = dom(y .--> x) by A2,RELAT_1:27;
  let e be object;
  assume
A3:  e in dom((A --> x)*(y .--> z));
  thus ((A --> x)*(y .--> z)).e = (A --> x).((y .--> z).e) by A3,FUNCT_1:12
      .= (A --> x).z by A3,Th7
      .= x by A1,Th7
      .= (y .--> x).e by A3,Th7;
 end;
