Logic / Discrete Mathematics
posted by Luke .
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.

Logic / Discrete Mathematics 
MathMate
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 
Logic / Discrete Mathematics 
Luke
@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.

Logic / Discrete Mathematics 
MathMate
You're welcome! :)
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I just cannot figure out if these answers are correct or not. Can you please help me?