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03-13-2013, 05:18 PM
  #118
Kane One
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Quote:
Originally Posted by Jabroni1994 View Post
Then use DeMorgan's Law and assoc. of OR:

*p OR (q OR r)
=== *p OR q OR r | by assoc. of OR
=== p AND *q AND *r | by DeMorgan's Law
Yep. That's what I did, but this is a ***** to check and see if I was right.

p --> (q OR r)
=== *p OR (q OR r)
=== **(*p OR (q OR r)
=== *(p AND *(q OR r))
=== *(p AND *q AND *r)

Looks right to me.

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