Understanding the p implies q statement

Question

The p implies q statement is often described in various ways including:
(1) if p then q (i.e. whenever p is true, q is true)
(2) p only if q (i.e. whenever q is false, p is false)

I see the truth table for (1) as

p | q | if p then q
-------------------
T | T | T
T | F |
F | T |
F | F |

I see the truth table for (2) as

p | q | p only if q
-------------------
T | T | 
T | F | F
F | T |
F | F | T

How are the two statements the same? What is wrong with my understanding?

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Addenda
1). There are some excellent excellent answers/suggestions here but what really worked for me was the following tip:

I think the intuitive way to think of this is if something is contradicted then it is false but if nothing can be contradicted it is by default true.

2). I have now learned that a conditional statement that is true by virtue of the fact that its hypothesis is false (i.e. true by default) is called vacuously true.

Answer 1

"If P then Q" means that whenever P is true, Q is true as you have observed. Thus, if both are true, it follows that the statement as a whole is true. Now, what if P is true, but Q is false? What does this imply about the statement as a whole? In this case, it is not true that "if P is true, Q is true". Hence, the statement as a whole is false.

Now, consider the cases where P is false. Does it matter whether Q is true or false? Again, we are evaluating the truthfulness of the statement that "If P is true, then Q is also." Since P is false, the statement no longer has a bearing on the factuality of the statement we are assessing. By convention, we thus assert that the statement is still true. (At the very least, it doesn't contradict the original statement. This is a complex topic; you would do well to look up some related questions regarding this. See this one for instance.)

Now, using the advice above, try to fill in the truth table for the other one. You will find that they are the same; therefore, the two are logically equivalent.

Answer 2

I think the intuitive way to think of this is if something is contradicted then it is false but if nothing can be contradicted it is by default true.

$p\to q$ means whenever $p$ is true we'll have $q$ true. So two rows of the table are

$\begin{matrix}p&q&p\to q\\T&T&T&\text{whenever p is true we must have q true... and we do}\\T&F&F&\text{whenever p is true we must have q true... and we don't}\end{matrix}$

But what of

$\begin{matrix}p&q&p\to q\\F&T&\\F&F\end{matrix}$

Well, in these cases our condition $p$ is true is not met so we are not obligated to anything about $q$. We can't contradict $p\to q$ as we can only test that if $p$ is true. And if these statements can't be contradicted.... they must be assumed to be true. So

$\begin{matrix}p&q&p\to q\\T&T&T&\text{whenever p is true we must have q true... and we do}\\T&F&F&\text{whenever p is true we must have q true... and we don't}\\F&T&T&\text{whenever p is true we must have q true, but we don't have p is true so all bets are off}\\F&F&T&\text{whenever p is true we must have q true, but we don't have p is true so all bets are off}\end{matrix}$

And the truth table for $q\text{ only if }p$ is similar

$\begin{matrix}p&q&q\text{ only if }p\\T&T&T&\text{whenever q is false we must have p true, but we don't have q is false so all bets are off}\\T&F&F&\text{whenever q is false we must have p false... and we don't}\\F&T&T&\text{whenever q is false we must have p true, but we don't have q is false so all bets are off}\\F&F&T&\text{whenever q is false we must have p false... and we do}\end{matrix}$

Same values.

Answer 3

Truth table last

First, on necessary and sufficient:

If A, then B. This means that anytime you have A, you will have B. It does not say that anytime you have B, you will have A. There could be times with B but not A. If anytime you have A you have B, then B is necessary for having A. That’s because you can’t have A without B. A is sufficient for B (knowing we have A, that is sufficient to know we have B).

These are all the same:

A $\implies$ B

If A, then B,

B $\impliedby$ A

B is necessary for A,

A is sufficient for B,

B if A

A only if B

There is no case with A but not B; there may be cases with B but not A,

A implies B,

In a venn diagram, the circle for the cases of A lies entirely within the circle for the cases of B.

Whenever A, then B ,

We can’t have A without B.

And....

A | B | statement

T | T | T

T | F | F

F | T | T

F | F | T

Meaning statement must be A implies B

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All those were the same

p | q | p only if q (this is p implies q)

T | T | T

T | F | F

F | T | T

F | F | T

These just means every incident of T,T did not violate (p implies q).

Whenever p and q are true (which is also written, whenever p and q), then we are at a point where $p \implies q$ is true. And also where $q \implies p$. What would violate $p \implies q$ and make it false? Well, if we have p but don’t have q.

Answer 4

$ p\implies q $ can be seen as an argument $ p \therefore q$. If the argument is valid and $ p $ is true, then we are sure that $ q $ is also true.

If the argument is valid and $ p $ is false, we can say nothing about $ q $.

using material implication means that we can replace $ p\implies q $ by $ \lnot p \vee q$.

Answer 5

Further to the other comments and answers that have dealt with your motivating query (my additional take here), I'd like to address your question and follow-up comment “I could be wrong but I don't believe the statements address the remaining entries in the truth table (the ones that I have left out)” at face value.

• Think of $\;\rightarrow\;$ (the material conditional) in symbolic logic as a logical operator—analogous to $\div$ being an arithmetic operator—that outputs a truth value depending on the truth values of its operands $P$ and $Q.$ Operators operate according to predefined rules; here, how $\;\rightarrow\;$ operates is actually defined by its truth table. The sentence $$P\rightarrow Q$$ (“if $P$ then $Q$”) is agnostic to the truth values of $P$ and $Q;$ it doesn't care whether its output is T or F.
• On the other hand, the assertion $$P\implies Q$$ (“$P$ implies $Q$”) encountered in non-formal logic (proofs or arguments) claims that the sentence ‘if $P$ then $Q$’ is true. It is asserting that $Q$ is a consequence of (can be concluded from) $P$ being true.

Answer 6

After practicing filling truth table and gaining logic terminologies, the natural language intuition for "if p then q" is generally that p is a sufficient condition of q, while for "p only if q" q is a necessary condition for p. With these intuitions you can usually find answers with more ease.

Final note about your natural language word "implies" which usually indicates logical or semantic consequence, and simple propositional logic you're learning here usually cannot fully express as explained further by my earlier post here.