theorem
  p is one-to-one & x in rng p implies {x..p} = p " {x}
proof
  assume that
A1: p is one-to-one and
A2: x in rng p;
  thus {x..p} c= p " {x}
  proof
    let y be object;
    assume y in {x..p};
    then y = x..p by TARSKI:def 1;
    hence thesis by A2,Th23;
  end;
  let y be object;
  assume
A3: y in p " {x};
  then
A4: y in dom p by FUNCT_1:def 7;
  p.y in {x} by A3,FUNCT_1:def 7;
  then
A5: p.y = x by TARSKI:def 1;
  p.(x..p) = x & x..p in dom p by A2,Th19,Th20;
  then x..p = y by A1,A4,A5;
  hence thesis by TARSKI:def 1;
end;
