# Biconditional statements with “or”

I have a question. Given a biconditional statement of the following form:

Show $p$ if and only if $q$ or $s$.

I was confused as to how to proceed.

In one direction of assuming $p$ is true, is it suffice to show $q$ is true? Or do I also need to consider if $p$ holds, then $s$ is true?

For the other direction, do I need to show two things: Assume $s$ is true and deduce $p$ and also assume $q$ and deduce that $p$ is true?

Thanks for the clarification.

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For the direction $p\to q\lor s$, you need to show that if $p$ is true, then at least one of $q$ and $s$ is true. If you can show that $p$ always implies $q$, then you don't need to worry about $s$ at all.

You are correct about how to prove the other direction, $(q\lor s)\to p$: show that $q\to p$ and $s\to p$. Alternatively, you can try to prove the contrapositive of $(q\lor s)\to p$, since it's logically equivalent: $\lnot p\to\lnot(q\lor s)$, or, using De Morgan's laws, $\lnot p\to(\lnot q\land\lnot s)$.

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Thanks, Brian. That helped me a lot. – MathNewbie May 4 '12 at 0:24

A common way to prove $p\to(q\lor s)$ is to convert $q\lor s$ to $\neg q\to s$ or $\neg s\to q$, and then prove $p\to(\neg q\to s)$, which is equivalent in turn to $(p\land \neg q)\to s$.

Informally, this argument is usually phrased by saying something like:

"Assume $p$; if $q$ holds, then we are done, so assume that $q$ does not hold. <stuff> hence $s$ holds, and we are done."

That is, you may assume both $p$ and the negation of one of $q$ and $s$, and prove the other from among $q$ and $s$. This is equivalent to proving $p\to(q\lor s)$.

Of course, if you can prove that $p\to q$, then since $q\to q\lor s$ is true, the two implications together give you $p\to(q\lor s)$.

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Thanks for the insight, Arturo. ;) – MathNewbie May 5 '12 at 6:39

Essentially, you need to show three things:

$$p \rightarrow q\text{ or }s$$ $$q \rightarrow p$$ $$s \rightarrow p$$

For the first one, if you like, you can equivalently show: $$p\text{ and ~}q \rightarrow s$$ or $$p\text{ and ~}s \rightarrow q$$ But you only need to show one of these two and ($p \rightarrow q\text{ or }s$), as they all imply each other.

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