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Math Logic
#10
Darkmaniak Wrote:I JUST started with this, and I can't do the exercises, I simply don't know where to start... I have to verify:
Code:
[p^(p=>q)]=>q
What I've got so far is
Code:
~[p^(p=>q)] v q
~p v ~(p=>q) v q
~p v (p ^ ~q) v q

After that step I just don't know what to do, also, just in case my symbols aren't used correctly, ^ conjunction, v as "or" and ~ as "no".
It should be universal language but I hope I didn't do anything stupid just cause I'm asking in english.

Kinda find hellen's transition from line 4 to 5 a bit iffy. Then again I haven't done this in a while. Found a less iffy way of doing it.

Code:
~[p^(p=>q)] v q
~[p^ (~p v q)] v q
~[ (p^~p) v (p^q)] v q  by distributivity law
~[ 0 v (p^q)] v q
~[ p^q] v q
(~p v ~q) v q
since these are all ORs, we can use associativity now

~p v (~q v q)
~p v 1  = 1

so the original statement is always true.
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Messages In This Thread
Math Logic - by Manu - 2011-03-08, 02:13 PM
Math Logic - by OB3LISK - 2011-03-08, 02:28 PM
Math Logic - by Shidoshi - 2011-03-08, 02:30 PM
Math Logic - by HellenzSin - 2011-03-08, 03:00 PM
Math Logic - by Kalovale - 2011-03-08, 03:15 PM
Math Logic - by Manu - 2011-03-08, 03:23 PM
Math Logic - by OB3LISK - 2011-03-08, 03:23 PM
Math Logic - by Kalovale - 2011-03-08, 03:32 PM
Math Logic - by HellenzSin - 2011-03-08, 03:49 PM
Math Logic - by shouri - 2011-03-08, 04:02 PM
Math Logic - by Manu - 2011-03-08, 04:11 PM
Math Logic - by HellenzSin - 2011-03-08, 04:12 PM
Math Logic - by Shidoshi - 2011-03-08, 04:15 PM
Math Logic - by HellenzSin - 2011-03-08, 04:49 PM

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