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If $p\implies q$ is the same as $\lnot p \lor q$, then what is $p\implies \lnot q$?

I'm not sure if this is $\lnot p \lor \lnot q$, or $\lnot p \lor q$.

I'm trying to figure this out, because i have a problem:

~(q v p) --> ~r). I use demorgans law on this to make it ~p ^ ~q --> ~r. Then I need to make it simpler, my understanding would be that it MIGHT be ~(~p ^ ~q) v ~r) which might be (p ^ q) v r. However, I'm almost 100% sure i'm wrong.

I can't wrap my head around logic, It's so difficult for me to comprehend all of these rules; could someone please explain the answer to my original question, and provide the correct answer for the ~(q v p) --> ~r) question; and where I went wrong? I would be very grateful.

p.s. if you know of any easy to comprehend resources for logic that would be great.

Thanks.

Thomas Andrews
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Anteara
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  • You are missing a parenthesis in ~(q v p) --> ~r). Does this mean ~((q v p) --> ~r or did you just mean ~(q v p) --> ~r? – Thomas Andrews May 17 '13 at 12:25
  • @ThomasAndrews well I have the problem; It is not true that if either [Bridget wins a silver medal] or [Carlos wins a gold medal] then [Janos does not win bronze medal]. I assign these as q, p, and ~r, in that order. So I understood this to mean ~(q v p) --> ~r... Is this wrong? – Anteara May 17 '13 at 12:37
  • I think that statement is: $\lnot\left((p\lor q)\implies \lnot r\right)$. The statement: $(p\lor q)\implies \lnot r$ is "If either p or q then not r." Then the entire statement has an "it is not true that..." before it. – Thomas Andrews May 17 '13 at 12:40
  • That makes sense, I guess I could split the above sentence up like this It is not true that (if either ([Bridget wins a silver medal] or [Carlos wins a gold medal]) then [Janos does not win bronze medal]), right? – Anteara May 17 '13 at 12:53

1 Answers1

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Well, $p \to r$ is the same as $\neg p \vee r$. Thus if $r=\neg q$, this means that $p \to \neg q$ is the same as $\neg p \vee (\neg q)$, which by order of logical operations can be written $\neg p \vee \neg q$.

No need to worry about De Morgan's laws!

Twiceler
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  • I believe I understand that there is no need for DeMorgans law in proving that p --> ~q is the same as ~p v ~q... (I even made a truth table to help :) - However, as I presently understand it when simplifying an expression it's best to apply all the laws you can and see where it goes from there... Can you show me the steps in simplifying ~(q v p) --> ~r)? Thanks. I do see a fallacy in using any law that's possible, since it would work both ways, possibly making it more complicated (did I make it more complicated by applying DeMorgans?) Thanks. – Anteara May 17 '13 at 12:40
  • Marking this as the accepted answer because i understand your first statement, and others seem to like your answer. However i don't understand why I don't need to worry about de morgan's laws when it comes to my second question. – Anteara May 17 '13 at 14:00