theorem Th54:
  p <==> q & r <==> s implies p '&' r <==> q '&' s
proof
  assume that
A1: p <==> q and
A2: r <==> s;
  s => r is valid by A2,Th50;
  then
A3: s => r in TAUT(A) by CQC_THE1:def 10;
  q => p is valid by A1,Th50;
  then q => p in TAUT(A) by CQC_THE1:def 10;
  then q '&' s => p '&' r in TAUT(A) by A3,PROCAL_1:56;
  then
A4: q '&' s => p '&' r is valid by CQC_THE1:def 10;
  r => s is valid by A2,Th50;
  then
A5: r => s in TAUT(A) by CQC_THE1:def 10;
  p => q is valid by A1,Th50;
  then p => q in TAUT(A) by CQC_THE1:def 10;
  then p '&' r => q '&' s in TAUT(A) by A5,PROCAL_1:56;
  then p '&' r => q '&' s is valid by CQC_THE1:def 10;
  hence thesis by A4,Th50;
end;
