Answer
the truth values of p* and q* are the same.
Work Step by Step
Let p and q be two compound propositions involving only the operators /\, V, and -
we can also allow them to involve the constants T and F.
We want to show that if p and q are logically equivalent, then p* and q* are logically equivalent.
The trick is to look at 'P and -.q.
They are certainly logically equivalent if p and q are.
Now if p is a conjunction, say r /\ s, then 'P is logically equivalent, by De Morgan's law, to -.r V -.s
A similar statement applies if p is a disjunction.
If r and/or s are themselves compound propositions, then we apply De Morgan's laws again to "push" the negation symbol -, deeper inside the formula, changing /\ to V and V to /\.
We repeat this process until all the negation signs have been "pushed in" as far as possible and are now attached to the atomic (i.e., not compound) propositions in the compound propositions p and q.
Call these atomic propositions p1 , p2 , etc.
Now in this process De Morgan's laws have forced us to change each /\ to V and each V to /\.
Furthermore, if there are any constants T or F in the propositions, then they will be changed to their opposite when the negation operation is applied: -.T is the same as F, and -.F is the same as T.
In summary, 'P and -.q look just like p* and q*, except that each atomic proposition p, within them is replaced by its negation.
Now we agreed that 'P = -.q; this means that for every possible assignment of truth values to the atomic propositions p1 , P2, etc., the truth values of 'P and -.q are the same. But assigning T to Pi is the same as assigning F to 'Pi , and assigning F to p, is the same as assigning T to 'Pi .
Thus, for every possible assignment of truth values to the atomic propositions, the truth values of p* and q* are the same. This is precisely what we wanted to prove.
