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.
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
@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.
You're welcome! :)
To show that "¬p Λ q" is logically equivalent to "p → ¬q", we can use a truth table or apply logical laws.
1. First, let's break down both propositions:
¬p Λ q: This is the conjunction (and) of the negation of p and q.
p → ¬q: This is the conditional statement (if-then) where p implies the negation of q.
2. Now, let's construct the truth table for both propositions:
p | q | ¬p | ¬q | ¬p Λ q | p → ¬q |
------------|--------------------|----------------|--------------
T | T | F | F | F | T |
T | F | F | T | F | T |
F | T | T | F | F | T |
F | F | T | T | T | T |
3. By comparing the truth values of ¬p Λ q and p → ¬q in the truth table, we can see that they have the same values for all possible combinations of truth values for p and q. Therefore, ¬p Λ q and p → ¬q are logically equivalent.
Alternatively, we can demonstrate the logical equivalence by applying logical laws. Here are the steps:
1. Start with the proposition ¬p Λ q.
2. Distribute the negation over p: (¬p) Λ q.
3. Use De Morgan's Law to move the negation inside the conjunction: ¬(p → ¬q).
4. Use the definition of the conditional statement to rewrite ¬(p → ¬q) as p → ¬q.
Thus, we have shown that ¬p Λ q is logically equivalent to p → ¬q using truth tables and logical laws.