OT Thread XIX: #NewYork****ingRangers

  • Thread starter Thread starter *Bob Richards*
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Dang it. Passed over again.

I don't think I'm ever going to be pope.

Don't feel bad, I missed out too. It's because they asked me how i felt about the Cardinals picking the right pope for the job and I answered with, "that's all well and good, but shouldn't they be more worried about their off season acquisitions?"
 
Don't feel bad, I missed out too. It's because they asked me how i felt about the Cardinals picking the right pope for the job and I answered with, "that's all well and good, but shouldn't they be more worried about their off season acquisitions?"
I think I lost points in the Q&A as well. Rocked it in the bathing suit competition though.
 
I think I lost points in the Q&A as well. Rocked it in the bathing suit competition though.

Too bad bathing suit comp is only 2 points. I won the majestic beard competition and Chris Hansen test, (ifyouknowwhatimean) But I guess they just felt this Francis guy was better.
 
For anyone who learned logic in school, can you tell me if this is correct?

Since I can't type the 'not' symbol, I'll use an asterisk instead.

p --> (q V r) <==> *(p ^ *q ^ *r)

p --> q <===> *p OR q.

So, p --> (q OR r) <===> *p OR (q OR r), where you instantiate q as (q OR r).
 
My professor asked us to rewrite it so that we only use 'not' and 'and'. We couldn't use 'or'.

Then use DeMorgan's Law and assoc. of OR:

*p OR (q OR r)
=== *p OR q OR r | by assoc. of OR
=== p ^ *q ^ *r | by DeMorgan's Law

And I got p --> q === *p OR q from the Discrete Mathematics and Its Applications (7e) by Rosen.
 
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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.
 
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.

Line 3 is missing a parenthesis and has an extra NOT at the beginning.

Other than that, it looks good. :nod:
 
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