theorem Th70:
  p in rng f implies rng f = rng(f-:p) \/ rng(f:-p)
proof
  assume
A1: p in rng f;
  then f-:p = (f -| p)^<*p*> by Th40;
  then
A2: rng( f-:p) = rng(f -| p) \/ rng<*p*> by FINSEQ_1:31;
  f:-p = <*p*>^(f |-- p) by A1,Th41;
  then
A3: rng(f:-p) = rng<*p*> \/ rng(f |-- p) by FINSEQ_1:31;
  f = (f-|p)^<* p *>^(f|--p) by A1,FINSEQ_4:51;
  hence rng f = rng((f-|p)^<* p *>) \/ rng(f|--p) by FINSEQ_1:31
    .= rng(f-|p) \/ (rng<* p *> \/ rng<* p *>) \/ rng(f|--p) by FINSEQ_1:31
    .= rng(f-|p) \/ rng<* p *> \/ rng<* p *> \/ rng(f|--p) by XBOOLE_1:4
    .= rng(f-:p) \/ rng(f:-p) by A2,A3,XBOOLE_1:4;
end;
