2011-03-08, 04:02 PM
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:
What I've got so far isCode:[p^(p=>q)]=>q
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.
