theorem Th60:
  Free(p => q) = Free p \/ Free q
proof
  p => q = 'not'(p '&' 'not' q) by QC_LANG2:def 2;
  hence Free(p => q) = Free(p '&' 'not' q) by Th39
    .= Free p \/ Free 'not' q by Th42
    .= Free p \/ Free q by Th39;
end;
