How do I determine equivalence between two forms of conditional statements?

Question

I'm reading Kenneth Rosen's Discrete Mathematics and its Applications. An excerpt for explaining how the truth table for "$p$ only if $q$" may be filled goes as follows:

“p only if q” says that p cannot be true when q is not true. That is, the statement is false if p is true, but q is false. When p is false, q may be either true or false, because the statement says nothing about the truth value of q.

We get the following truth table from that justification. $$\begin{array}{|c|c|c|} \hline p&q&p\Rightarrow q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T\\\hline \end{array}$$

My confusion is related to "$q$ if $p$" resulting in the same table As far as I understand, "$q$ if $p$" implies $q$ can be true if $p$ is true or even if it isn't, which would fill the truth table like so:

$$\begin{array}{|c|c|c|} \hline p&q&p\Rightarrow q\\ \hline T&T&T\\ F&T&T\\\hline \end{array}$$

Following Rosen's When p is false, q may be either true or false, because the statement says nothing about the truth value of q. Can I say when $q$ is false, $p$ may either be true or false, because the statement says nothing about the truth value of $p$? But that results in the wrong truth table, can someone explain what I'm missing here?

Answer 1

Short answer : " Q, if P" would be false in case ...

You may just complete the sentence and put the truth-value " F" on the corresponding row(s) of the truth-table. Put " T" on all other rows, for whenever a sentence is not false, it is true, by default.

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My confusion is related to "q if p" resulting in the same table As far as I understand, "q if p" implies q can be true if p is true or even if it isn't, which would fill the truth table like so:

\begin{array}{|c|c|c|} \hline p&q&p\Rightarrow q\\ \hline T&T&T\\ F&T&T\\\hline \end{array}

• What you say is perfectly correct, but the conclusion you draw in terms of truth table is wrong.

A principle that can be helpful here is :

whenever the falsity conditions of a sentence are not met, this sentence is true, by default.

• So, let's imagine John asserts " Q, if P", and let's find in which case John is wrong.

The only case in which we would be entitled to say that John is wrong is the case where P is true, and where, in spite of that, Q is not true.

Saying " Q, if P" is the same as saying " if it happens that Q is false while P is true, I am wrong "

Only the occurrence of such a case would show John was wrong in asserting that Q is always true when P is true, or, if you prefer, that " P is a sufficient conditon for Q to be true" ( For john didn't say more than this).

Since the case ( T,F) is the only falsity condition, the conditional is true in all other possible cases.

• This results in the ordinary truth table of the " if...then " operator.

Answer 2

No, you cannot say that, because the relation is asymmetric. You're confusing $p\implies q$ with its converse $q\implies p$. It's actually equivalent to its contrapositive $\neg q\implies\neg p$. The reason $p\implies q$ is true when $p$ is false, but not necessarily if $q$ is false, is because it's effectively saying $q$ is at least as true as $p$. Writing false as $0$ and true as $1$, $\implies$ becomes $\le$.

Answer 3

Think of $p \Rightarrow q$ as a standard theorem. For example, "If x is 2, then x is even."

Let $p$ be the part "x is 2" and $q$ be the part "x is even." Now, say x is not even. Can it be 2? No. This would be the contrapositive argument $\neg q \Rightarrow \neg p$ which has an equivalent truth table. So, if $q$ false, $p$ false.

Now say x is not 2 ($\neg p$). Can x be even? Yes. So, we can have that $q$ is true (though it is not necessarily true). Similarly, if x is even ($q$ holds), x may or may not be 2, so $p$ may or may not hold.

When dealing with implication, we care less about when the implication is true and more about when it is false. It is only false when $p$ is true, but $q$ is not. i.e. if "x is 2" then "x is not even" -- obviously false!

Answer 4

My confusion is related to "$q$ if $p$" ...

Following Rosen's When p is false, q may be either true or false, because the statement says nothing about the truth value of q. Can I say when $q$ is false, $p$ may either be true or false, because the statement says nothing about the truth value of $p$?

No, you can't say that, because given the statement "$q$ if $p$", then it follows that if $q$ is false, $p$ will have to be false as well. Consider, suppose that $p$ were true. Then given the statement "$q$ if $p$", $q$ itself would have to be true as well. So, if $q$ is in fact false, that means $p$ cannot be true, and will therefore have to be false as well.