show that ¬(p →¬q) is logically equivalent to (p∧ q)

Question

¬(p →¬q)≡ q∧p Commutative Law ≡ p∧q Commutative Law

That's my solution. Is it right?

Answer 1

p->q='p V q

In your case p->'q=p'V q'

When you invert it: '(p' V q')=p ∧ q

This is the final answer.

Answer 2

If your question is: is the final answer right, then yes your answer is correct, however your justification is incorrect. You need a little more to get there: $\neg (p \rightarrow \neg q)$ can be written as $\neg (\neg p \lor \neg q)$ by the definition of $\rightarrow$ and applying DeMorgan's law we get that $\neg (p \rightarrow \neg q)$ is the same as $p \land q$

Answer 3

It is clear when $p \land q$ is true: if, and only if, both $p$ and $q$ are true.

When is $\neg(p \rightarrow \neg q)$ true? Well, when the parenthetical $p \rightarrow \neg q$ is false. Noting that this equivalent statement is an if-then implication, it can only be false if the given is true and the conclusion is false; i.e., if $p$ is true and $\neg q$ is false, from which the latter can be rephrased as $p$ true and $q$ true: precisely the desired conclusion.

Answer 4

¬(p->¬q) ≡ ¬(¬p∨¬q)
¬(p->¬q) ≡ p∧q

Let's try to visualise the problem. This can be explained with ven diagram also. (p->¬q) means, if p is true, then q is not true for some x in any of the 4 faces of the ven diagram with p and q. Let's color the area's with white if x satisfies p->¬q. else black.

1. In the area p-q, p is true, q is not true. So color will be white as p->¬q is satisfied.
2. In the area q-p, p is false. So p->¬q will be automatically satisfied. As the case when p is false is not specified in p->¬q. So we color this area with white.
3. In the area U-(q∪p), again p is not true. So, this area will also be white.
4. In the are p∩q, p is true and q is also true. Which is false as we have said, if p is true, q muust be false. So, we color this area black.

Our final argument says to negate(p->¬q). So, only p∩q will be white and all other parts of the ven diagram will be false, which proves that ¬(p->¬q) implies p∩q or, p∧q.