up vote 0 down vote favorite
share [g+] share [fb]

This is for homework for my Discrete Math class.

I understand how to prove tautologies with a truth table. I'm having trouble figuring out how to do it without though.

I found this example in our text, but the solution is not provided. I was hoping someone could answer it as best they could so I could apply it to my other questions. Thanks!

! = not/invert
^ = and
V = or
-> = if then

[!P ^ (P V Q) ] -> Q

I also am completely lost for this question (I fear I'll be tested on something of this difficulty)

[ (P -> Q) ^ (Q -> R) ] -> (P -> R)

link|improve this question
as much as i love math personally, this is not the kind of question this site is here for. you might try mathoverflow.com but i don't know how welcoming they are of homework questions. – quack quixote Mar 22 '10 at 23:58
Alright, i thought it may be appropriate because it is computer-related math. I am a compsci major, so i thought other compsci students/graduates may visit these boards. I'll give mathoverflow a try. – user31996 Mar 23 '10 at 0:09
2  
Mathoverflow is for research math level questions. I don’t think, you’re much better there. – Debilski Mar 23 '10 at 0:12
1  
Superuser isn't here to do your homework. – Josh K Mar 23 '10 at 1:26
feedback

closed as off topic by quack quixote, heavyd, William Hilsum, hyperslug, Josh K Mar 23 '10 at 1:26

Questions on Super User are expected to generally relate to computer software or computer hardware, within the scope defined in the faq.

1 Answer

up vote 1 down vote

First problem: Use distributivity and that P->Q is !P v Q.

Second problem: Expand ->, De-Morgan, distributivity, try to apply the last (!P v R) to all pairs of (X v Y) on the left.

link|improve this answer
feedback