Originally Posted by
poutsos.B To have that (A-B)' IS equal to B'-A' THIS must hold for all A AND B.
However if this equality does not hold then there must be an A or B or both such that :Let B= Φ ( the empty set), and if substitute Φ ΙΝ (Α-Β)' we have:
( A-Φ)'= (A)'= A' BECAUSE A-Φ=Α.Αnd substituting Φ in B'-A' we have:
Φ'-A' = U- A' =A BECAUSE Φ'=U WHERE U is the set of which A, B are subsets.
Hence we have A'= A NOT true.
But we can give a proof in propositional calculus without using a counter example by the following procedure:
Let xε( A-B)' <===> ~xεΑ v xεB as it was shown
Let xε(B'-A') <===> ~xεB & ~xεA' <====> ~xεB & xεA.
Now put xεA=p and xεB=q. So if we want (A-B)' = B'-A' W must have ,by the axiom of extensionality :
xε(A-B)' <====> xε(B'-A') or ~xεΑ v xεB <=====>~xεB & xεA or
~p v q <=====> p & ~q which means that ~p vq logicaly implies p & ~q and p & ~q logicaly implies ~p v q,which means that ;
The conditional ~p v q ------> p & ~q must be a tautology and,
The conditional p & ~q ------->~p v q must also be a tautology.
Now by forming the true tables of the above we see that none of the above is a tautology.
Now by a theorem in propositional logic that states:
A formula is a theorem ( hence provable ) of the propositional logic if and only if it is a tautology.
And the fact that the above two conditionals are not tautologies we conclude:
~p v q <=====> p & ~q is not provable ,hence xε(A-B)' <====> xε(B'-A') is not provable and ,
(A-B)' Is not equal to B'-A'
Note ~xεΑ x does not belong to A, AND ~p means not p