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April 19, 2015

Posted by **rick** on Tuesday, October 5, 2010 at 2:01am.

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!

- math -
**MathMate**, Tuesday, October 5, 2010 at 3:48pmYou'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.

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