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!
You'll need the logical equivalence:
p → q ≡ !p ∨ q
so
!p → (q→r)
≡ p ∨ (!q ∨ r)
From here, use the commutative properties to rearrange the expression and apply the equivalence (of →) again to get the desired result.
To show that the propositions !p -> (q->r) and q -> (p v r) are logically equivalent, we need to use logical equivalences to transform one proposition into the other.
Starting with !p -> (q->r), we can apply the implication rule:
!p -> (q->r)
= !p -> (!q v r)
Now, let's focus on transforming q -> (p v r). We can start with the implication:
q -> (p v r)
To transform this, we can use the rule of material implication:
= !q v (p v r)
Now, let's compare the two propositions:
!p -> (!q v r)
!q v (p v r)
We can see that these two propositions have the same structure. To show that they are logically equivalent, we can apply the commutative property to swap the order of the terms in the second proposition:
!p -> (!q v r)
!q v (r v p)
Now, we can notice that the propositions !q v (r v p) and !q v (p v r) are identical. This can be shown by applying the associative property:
!q v (r v p)
= (!q v r) v p
Since !q v r is logically equivalent to !q v (p v r), we can conclude that the propositions !p -> (q->r) and q -> (p v r) are logically equivalent.