reserve A for QC-alphabet;
reserve i,j,k for Nat;
reserve f for Substitution of A;

theorem Th12:
  for r,s being Element of CQC-WFF(A) holds
  r <=> s is Element of CQC-WFF(A)
proof
  let r,s be Element of CQC-WFF(A);
  r <=> s = (r => s) '&' (s => r) by QC_LANG2:def 4
    .= ( 'not' (r '&' 'not' s) ) '&' (s => r) by QC_LANG2:def 2
    .= ( 'not' (r '&' 'not' s) ) '&' ( 'not' (s '&' 'not' r) ) by
QC_LANG2:def 2;
  hence thesis;
end;
