I want to prove or disprove (A-B)'=B'-A'.
So, let x be an element of (A-B)'
Then x is not an element of A-B
This is where I get confused...
Do I say x is an element of A', x is an element of -B'?
I agree that that is not very helpful!
Mathematicians usually hate such proofs done with logic.
You ought to know that the statement is formal false.
Counterexample:
$\displaystyle \begin{gathered}
U = \left\{ {1,2,3,4,5} \right\},\,A = \left\{ {1,3,4,5} \right\}\,\& \,B = \left\{ {2,3,5} \right\} \hfill \\
\left( {A\backslash B} \right)^\prime = \left( {\left\{ {1,4} \right\}} \right)^\prime = \left\{ {2,3,5} \right\} \hfill \\
B'\backslash A' = B' \cap A = \left\{ {1,4} \right\} \hfill \\
\end{gathered} $
BTW: We use this $\displaystyle \left( {A - B} \right) = \left( {A\backslash B} \right)$ it is called "setminus".
i assure you, that is not what your teacher taught. as Plato said, to disprove something, all you have to do is give a counter-example. if you can find one instance where a claim is false, then the claim is considered false in general. anything else is just giving yourself ideas as to why something might be false. as you said, you drew Venn Diagrams and found that the claim might not be true. so then you know you should start looking for a counter-example. you could also notice that the left hand side, using the defintion of set minus and DeMorgan's laws, simplifies to $\displaystyle \neg A \cup B$, which is not equal to the other side.
these are not proofs that the claim is false, however, to disprove something, you must give a counter example
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 :
(A-B)' IS not equal to B'-A' AND Plato already gave you an example.
Perhaps an example not using Nos is the following:
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
All of that is truly preposterous. Only a logician could have produced that.
For a mathematical statement it is totally meaningless.
I think that is a perfect example why “Symbolic Logic” is dieing as a separate subject.
The American Symbolic Logic Society meets at the joint meeting of the American Mathematical Society and The Mathematical Association of America. Over the last thirty years I have tried to attend sessions of the Logical Association, but lately the average session has no more that five in attendance