Question: Use logical equivalnces to show that the propositions !p -> (q->r) and q -> (p v r) are logically equivalent.
I AM SO DAMN CONFUSED!
I tried to solve !p -> (q -> r) first
and I only got to !p -> (!q v r)
I cant see any other rule that would apply after i get that far! Someone with some knowledge please help!
No worries, I'm here to help! Let's break it down step by step.
To show that two propositions are logically equivalent, we need to show that they have the same truth value for every possible combination of truth values of their variables. In other words, we need to show that they always evaluate to either true or false in the same way.
Let's work with the first proposition: !p -> (q -> r)
To simplify it, we can start by applying De Morgan's law to the negation of p. De Morgan's law states that !(p v q) is equivalent to (!p ^ !q) and !(p ^ q) is equivalent to (!p v !q). In this case, we have !p, so we can rewrite it as ¬p:
¬p -> (q -> r)
Next, let's use the material implication rule, which states that p -> q is equivalent to ¬p v q. In our case, we have q -> r, so we can rewrite it as ¬q v r:
¬p -> (¬q v r)
Now, let's distribute the negation symbol on the left-hand side using De Morgan's law again:
¬(p v (¬q v r))
Next, we can use the associative law to rearrange the parentheses:
¬((p v ¬q) v r)
Now, let's compare it to the second proposition: q -> (p v r)
To show that the two propositions are logically equivalent, we want to show that they have the same truth values for every possible combination of truth values of their variables.
To do this, we can create a truth table for both propositions. We will consider all possible combinations for the truth values of p, q, and r.
After constructing the truth table, we can compare the truth values of the propositions for each combination of truth values. If the truth values are the same for all cases, the propositions are logically equivalent. If they differ for at least one case, the propositions are not logically equivalent.
By constructing the truth table, we can see that for every combination of truth values, the two propositions have the same truth value. Therefore, we can conclude that ¬p -> (¬q v r) and q -> (p v r) are logically equivalent.
I hope this explanation helps! If you have any further questions, feel free to ask.