Quote:
Originally Posted by TheDon
its helping.. a little.. but I just cant seem to still get it.
Prove (p → r) ∨ (q → r) ≡ (p ∧ q) → r
but since there are no ~ De Morgans law wont come into play here.. I think.
So I have to make the right equal to the left. so I can leave the left alone, correct?
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DeMorgans comes into play later down in the problem.
You have to re-write p->r and q->r into basic and/or form. See my post a few back.
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