theorem Th2:
  p => (q => r) is valid implies (q '&' p) => r is valid
proof
  assume p => (q => r) in TAUT(A);
  then p => (q => r) is valid;
  then (p '&' q) => r is valid by Th1;
  then
A1: (p '&' q) => r in TAUT(A);
  q '&' p => p '&' q in TAUT(A) by CQC_THE1:45;
  hence (q '&' p) => r in TAUT(A) by A1,LUKASI_1:3;
end;
