reserve x,y,z for set;
reserve f,f1,f2,f3 for FinSequence,
  p,p1,p2,p3 for set,
  i,k for Nat;
reserve D for non empty set,
  p,p1,p2,p3 for Element of D,
  f,f1,f2 for FinSequence of D;

theorem Th71:
  p1 in rng f & p2 in rng f \ rng(f-:p1) implies f:-p1:-p2 = f:-p2
proof
  assume that
A1: p1 in rng f and
A2: p2 in rng f \ rng(f-:p1);
A3: not p2 in rng(f-:p1) by A2,XBOOLE_0:def 5;
  f-:p1 = (f -| p1)^<*p1*> by A1,Th40;
  then rng(f-:p1) = rng(f -| p1) \/ rng<*p1*> by FINSEQ_1:31;
  then
A4: not p2 in rng<*p1*> by A3,XBOOLE_0:def 3;
  rng f = rng(f-:p1) \/ rng(f:-p1) by A1,Th70;
  then
A5: p2 in rng(f:-p1) by A2,A3,XBOOLE_0:def 3;
  f:-p1 = <*p1*>^(f |-- p1) by A1,Th41;
  then rng(f:-p1) = rng<*p1*> \/ rng(f |-- p1) by FINSEQ_1:31;
  then p2 in rng(f|--p1) by A5,A4,XBOOLE_0:def 3;
  then
A6: p2 in rng(f|--p1) \ rng<*p1*> by A4,XBOOLE_0:def 5;
  thus f:-p1:-p2 = <*p2*>^((f:-p1)|--p2) by A5,Th41
    .= <*p2*>^((<*p1*>^(f|--p1))|--p2) by A1,Th41
    .= <*p2*>^(f|--p1|--p2) by A6,Th9
    .= <*p2*>^(f|--p2) by A1,A2,Th69
    .= f:-p2 by A2,Th41;
end;
