theorem
  p <==> q & r <==> s implies p <=> r <==> q <=> s
proof
  assume that
A1: p <==> q and
A2: r <==> s;
A3: r => p <==> s => q by A1,A2,Th55;
A4: p <=> r = (p => r) '&' (r => p) by QC_LANG2:def 4;
  p => r <==> q => s by A1,A2,Th55;
  then (p => r) '&' (r => p) <==> (q => s) '&' (s => q) by A3,Th54;
  hence thesis by A4,QC_LANG2:def 4;
end;
