theorem Th65:
  p in rng f2 \ rng f1 implies (f1^f2):-p = f2:-p
proof
  assume
A1: p in rng f2 \ rng f1;
  rng(f1^f2) = rng f1 \/ rng f2 by FINSEQ_1:31;
  then p in rng(f1^f2) by A1,XBOOLE_0:def 3;
  hence (f1^f2):-p = <*p*>^((f1^f2)|--p) by Th41
    .= <*p*>^(f2|--p) by A1,Th9
    .= f2:-p by A1,Th41;
end;
