Help with Discrete Math homework
So I'm supposed to simplify the following statement, but I'm not sure if where I finished off is the final simplified version or if there's a more simplified version of it:
not ( p ^ ( q v r ) ^ ( ( p^ q ) -> r ) )
= not ( p ^ ( q v r ) ^ ( p -> r ) Rule of Simplification
= not ( ( p ^ q ) v ( p ^ r ) ^ ( p -> r ) ) Distributive laws
= not ( p v ( p ^ r ) ^ ( p -> r ) ) Rule of Simplification
= (not sure about this step) not p ^ not (p ^ r ) v not ( p -> r ) Multiplied not by everything inside
I just started this class so I have no idea if what I'm doing is right or wrong. xD Thanks in advance.
September 26, 2013
8 Comments • Newest first
[quote=Powerbomb]I think not ( p -> r ) gets negated to (not p ^ not r)[/quote]
it's actually p ^ not r
proof: not ( p -> r )
= not (not p v r) (Implication)
= p ^ not r (De Morgan's)
edit:
This should be everything you need to know to simplify the expression:
p -> q <=> ~p v q (implication)
~(p^q) <=> ~p v ~q (De Morgan's)
~(pvq) <=> ~p ^ ~q (De Morgan's)
p^(qvr) => (p^q) v (p^r) (Distribution)
pv(q^r) => (pvq) ^ (pvr) (Distribution)
pv(p^q) => p (Absorption)
p^(pvq) => p (Absorption)
@Blackyoshi: discrete math is hard, don't worry. The language is different and everything and I'm finding it to be harder than my other math classes
[quote=Powerbomb]I think not ( p -> r ) gets negated to (not p ^ not r).
You might want to negate the whole thing first and then simplify it. Your second statement looks strange, too. for (p -> r), where did q go?[/quote]
Ok thanks, I'll try that approach! I tried using the Rule of Simplification which says p^q, therefore p, with the corresponding tautology (p^q) -> p. And then I thought it was ok if I could just substitute p^q for p and then make it (p -> r)...I don't know what I'm doing LOL.
I think not ( p -> r ) gets negated to (not p ^ not r).
You might want to negate the whole thing first and then simplify it. Your second statement looks strange, too. for (p -> r), where did q go?
Time to ditch and go into software engineering.
I think you can simplify it more if you substitute (not p v r) for the (p -> r). I don't remember much from this class though.
[quote=HolyDragon]Is this a math major course or something?[/quote]
No, it's for computer science. If I was a math major, I'd die D:
Is this a math major course or something?