theorem Th66:
  p in rng f1 implies (f1^f2)-:p = f1-:p
proof
  assume
A1: p in 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 = ((f1^f2)-|p)^<*p*> by Th40
    .= (f1-|p)^<*p*> by A1,Th12
    .= f1-:p by A1,Th40;
end;
