theorem Th12:
  r <> 0 implies (r(#)f)"{0} = f"{0}
proof
  assume
A1: r <> 0;
  now
    let c;
    thus c in (r(#)f)"{0c} implies c in f"{0c}
    proof
      assume
A2:   c in (r(#)f)"{0c};
      then
A3:   c in dom (r(#)f) by PARTFUN2:26;
      (r(#)f)/.c in {0c} by A2,PARTFUN2:26;
      then (r(#)f)/.c = 0c by TARSKI:def 1;
      then r*(f/.c) = 0c by A3,Th4;
      then (f/.c) = 0c by A1,XCMPLX_1:6;
      then
A4:   (f/.c) in {0c} by TARSKI:def 1;
      c in dom f by A3,Th4;
      hence thesis by A4,PARTFUN2:26;
    end;
    assume
A5: c in (f)"{0c};
    then (f/.c) in {0c} by PARTFUN2:26;
    then (f/.c) = 0c by TARSKI:def 1;
    then
A6: r*(f/.c) = 0c;
A7: c in dom f by A5,PARTFUN2:26;
    then c in dom (r(#)f) by Th4;
    then (r(#)f)/.c = 0c by A6,Th4;
    then
A8: (r(#)f)/.c in {0c} by TARSKI:def 1;
    c in dom (r(#)f) by A7,Th4;
    hence c in (r(#)f)"{0c} by A8,PARTFUN2:26;
  end;
  hence thesis by SUBSET_1:3;
end;
