Show that "¬p Λ q is logically equivalent to p → ¬q" . I'm not sure if the conditional identity is correct, but what I did was I negated the whole second condition and it immediately became equal to the first? Please help and if possible, show steps or what laws I have to use. Thank you.

Answer 1

With simple expressions like these, the easiest proof is to make a truth table...unless it is specifically prohibited in the question.

Also, there are precedence in the logical operators, just like + - * ÷.
see
http://en.wikipedia.org/wiki/Logical_connective

The ¬ operator (which I write as ~ for simplicity) has the highest priority, in this order.

Operator Precedence
¬ 1
∧ 2
∨ 3
→ 4
↔ 5

There the first expression is interpreted as (~p)∧q which is not an identity to p->~q.

However, writing the first expression as ~(p∧q) will give a truth table of
TT F
TF T
FT T
FF T
identical to that of
p->~q
TT F
TF T
FT T
FF T
hence the identity
~(p∧q) ≡ p->~q

Answer 2

@MathMate Thank you so much! No wonder I was going nowhere even when I did the truth table, I had a mistake in translating the proposition. Thanks again.

Answer 3

You're welcome! :)