In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True?

Question

Provided we have this truth table where "$p\implies q$" means "if $p$ then $q$":

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

My understanding is that "$p\implies q$" means "when there is $p$, there is q". The second row in the truth table where $p$ is true and $q$ is false would then contradict "$p\implies q$" because there is no $q$ when $p$ is present.

Why then, does the third row of the truth table not contradict "$p\implies q$"? If $q$ is true when $p$ is false, then $p$ is not a condition of $q$.

I have not taken any logic class so please explain it in layman's terms.

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Administrative note. You may experience being directed here even though your question was actually about line 4 of the truth table instead. In that case, see the companion question In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False? And even if your original worry was about line 3, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.

Answer 1

If you don't put any money into the soda-pop machine, and it gives you a bottle of soda anyway, do you have grounds for complaint? Has it violated the principle, "if you put money in, then a soda comes out"? I wouldn't think you have grounds for complaint. If the machine gives a soda to every passerby, then it is still obeying the principle that if one puts money in, one gets a soda out.

Similarly, the only grounds for complaint against $p\to q$ is the situation where $p$ is true, but $q$ is false. This is why the only F entry in the truth table occurs in this row.

If you imagine putting an F on the row to which you refer, the truth table becomes the same as what you would expect for $p\iff q$, but we don't expect that "if p, then q" has the same meaning as "p if and only if q".

Answer 2

$p\Rightarrow q$ is an assertion that says something about situations where $p$ is true, namely that if we find ourselves in a world where $p$ is true, then $q$ will be true (or otherwise $p\Rightarrow q$ lied to us).

However, if we find ourselves in a world where $p$ is false, then it turns out that $p\Rightarrow q$ did not actually promise us anything. Therefore it can't possibly have lied to us -- you could complain about it being irrelevant in that situation, but that doesn't make it false. It has delivered everything it promised, because it turned out that it actually promised nothing.

As an everyday example, it is true that "If John jumps into a lake, then John will get wet". The truth of this is not affected by the fact that there are other ways to get wet. If, on investigating, we discover that John didn't jump in to the lake, but merely stood in the rain and now is wet, that doesn't mean that it is no longer true that people who jump into lakes get wet.

However, one should note that these arguments are ultimately not the reason why $\Rightarrow$ has the truth table it has. The real reason is because that truth table is the definition of $\Rightarrow$. Expressing $p\Rightarrow q$ as "If $p$, then $q$" is not a definition of $\Rightarrow$, but an explanation of how the words "if" and "then" are used by mathematicians, given that one already knows how $\Rightarrow$ works. The intuitive explanations are supposed to convince you (or not) that it is reasonable to use those two English words to speak about logical implication, not that logical implication ought to work that way in the first place.

Answer 3

From the other answers, the most convincing and reasonable explanation why logic implication is defined the way it is, is the idea of sufficient (versus "necessary") condition for something to be true.

NOTE: I don't buy the answer/argument that "if-then" is not an equivalent definition of "$\implies$"; it is just that we tend to have a different notion of "if-then" in everyday life, that of a necessary one.

In short:

Any if-then statement that would "break" only when the conclusion is false while the condition is true is a logic implication.

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Examples from everyday life

The following are examples of statements that are logical implications.

1. "if (it is raining), then (there are clouds in the sky)"

   • $T$ - $T$ (OK checks out) $T$
   • $T$ - $F$ (Oops!) $F$
   • $F$ - $T$ (OK, there are clouds in the sky but it doesn't rain, but that does not "break the rule") $T$
   • $F$ - $F$ (no rain, no clouds, still a valid statement as far as I can tell) $T$

2. "if (I find my room not the way I left it), then (someone was in my room)"

   • $T$ - $T$ (ok checks out so far) $T$
   • $T$ - $F$ ("rule" breaks) F
   • $F$ - $T$ (does not break the logic of the "rule") $T$
   • $F$ - $F$ (this does not break the logic either) $T$

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A more elaborate explanation with example from science

Consider an example in the field of medical diagnosis. The basic (and ideal) premise of diagnosis from symptoms is to derive valid, sufficient rules that can safely conclude a diagnosis of an illness over other illnesses based on symptom observations. Let's say some medical scientist studies illness A and proposes the following diagnostic rule:

"IF (symptom B and symptom C are observed) THEN (--for sure-- the patient is under illness A)."

He then goes through all the documented cases of the illness (or conducts a new study) and tries to see whether that rule is valid:

1. If a patient in the records had the symptoms and he was also found to have illness A (1st row of the truth table), then so far so good.

2. If a patient is found to have the symptoms but not the illness, that breaks or falsifies the rule (2nd row in the truth table), and the rule has to be reconsidered and revised because it simply does not work; the rule, as a logic implication, is false.

3. If some patient is found to have illness C but not the symptoms (3rd row in the truth table), that does not reduce the validity of the rule in any way as a way of making a safe conclusion; it only reduces its usefulness, depending on how many cases it may miss.

4. If some patient is found not to have the symptoms nor the illness (4th row in the truth table), that is irrelevant to the validity of the rule.

So, if the scientist finds only records of the 1st, 3rd, and 4th case, then he has a valid rule. Moreover, the potential of the rule to break in the 2nd case makes it a logic implication.

From the above, you can see that the way logic implication is defined (with the third and fourth rows being True) finds extensive use in math and science and, eventually, is what makes sense.

Answer 4

To understand why this table is the way it is, consider the following example:

"If you get an A, then I'll give you a dollar."

The statement will be true if I keep my promise and false if I don't.

Suppose it's true that you get an A and it's true that I give you a dollar. Since I kept my promise, the implication is true. This corresponds to the first line in the table.

Suppose it's true that you get an A but it's false that I give you a dollar. Since I didn't keep my promise, the implication is false. This corresponds to the second line in the table.

What if it's false that you get an A? Whether or not I give you a dollar, I haven't broken my promise. Thus, the implication can't be false, so (since this is a two-valued logic) it must be true. This explains the last two lines of the table.

@attribution: http://www.millersville.edu/~bikenaga/math-proof/truth-tables/truth-tables.html

Answer 5

The statement $(P \land Q) \to P$ should be true, no matter what. So, we should have:

\begin{array}{cc|ccc} P&Q&(P \land Q) & \to & P\\ \hline T&T&T&T&T\\ T&F&F&T&T\\ F&T&F&T&F\\ F&F&F&T&F\\ \end{array}

Line 2 shows that we should therefore have that $F \to T = T$

Also note that line 1 forces $T \to T = T$, and that line 4 forces $F \to F=T$, which are another two values of the truth-table for $\to$ that people sometimes wonder about. So, together with the uncontroversial $T \to F = F$, the above give a justification for why we define the $\to$ the way we do.

Answer 6

• If a dog, then has 4 legs - true
• If a dog, then has no legs - false
• If not a dog (may be a cat), then has 4 legs - true
• If not a dog (may be a snake), then has no legs - true

Answer 7

The article on implication written by Timothy Gowers in his blog should be a nice (and helpful) reference to have here.

Answer 8

1. The fact that the conditional $$P\to Q$$ is true whenever its antecedent $P$ is false (principle of explosion; vacuous truth) is actually so by definition:

         $P\to Q\,$ is a truth function that is tautologically equivalent to $\,\lnot P\lor Q.$

   So, $P\to Q\,$ is false precisely when $P$ is true but $Q$ false.

2. To be clear: whenever $P$ is false, the assertion $P{\implies} Q\,,$ while true, gives no information about whether $Q$ is true.

3. Summarising these two explanations of the motivation for the above definition:

   if we insist, to the contrary, that  False$\to$True  be false, then, unfortunately, these violations of natural deduction arise: $$\text{$A$ is true and $B$ is false $\implies\Big[(A\land B)\to A\Big]\;$is false!}$$ and $$\Big[\forall n\in\mathbb Z \;\big(n \text{ is a multiple of }4\, \to \,n \text{ is even}\big)\Big]\;\text{is false}!$$

4. It is worth noting that in logic/mathematics, $P$ need not cause $Q$ for $P$ to imply $Q,$ that is, for the material conditional $\,P\to Q\,$ to actually be true.

   After all, the logical connective $\,\to,$ being a truth-functional operator, cares about truth states without considering the flow of time.

Answer 9

P implies Q means that Q is true when ever P is; it does not mean in addition that Q is false when ever P is...otherwise as a net result Q will equate to P.........No.

The sentiment here is causation and in that : P is a sufficient condition for Q and there may be other as well.

Therefore when P is false Q can be both true and false in the truth table (where such entries are accepted as true) the exact value of Q depending on other sufficient conditions.

Answer 10

I spent a lot of time thinking about it and looking for answers. I think I found it for myself. I will share it to verify its usefulness to others.

Working in any profession we need to verify what is true and what is not. We need the tools to do it. One uses a magnifying glass, another a French key, and another tool is this table. I look at the table as a whole. I don't see the arrow or the rows. I see the tool.

Let's go to the pharmacy. We have a dresser in the pharmacy. The dresser has small and large drawers. There are 100 drawers. 50 small and 50 big. You are an inspector. Your task for today is to make sure that:

• the children's medicines are only in the small drawers.

From now on, the table is a tool for me to check this problem. The baby drug detector in the not small drawers The tool has three green T lights and one red F light. When the red one lights up it means we have detected something we are looking for.

We want to turn on the fourth light, the read F one. How do we do that?

Checking the small drawers will not answer the question. We need to check the large drawers. We just need to find one children's medicine in the big drawer.

Translating the problem into tables we have:

P - children's medicine

Q - small drawer

We are looking for a row in the table in which there is: Children's medicine T and large drawer F. This notation tells us that its occurrence contradicts the assumption that in a dresser children's medicine is only in small drawers.

But what about F-> T ? Well, you just found a fork in the small drawer. Does that prove anything? Perhaps in another matter, yes. In this one, no.

To know that there are only black swans in the world, one looks for a swan of a different color. Not a white swan. A world with only white swans is a belief that the world is like that. Finding a black swan is finding a T -> F which tells us that this is not a world of only white swans.

The table must be seen as a whole in the context of considering a problem.

Answer 11

Every logical statement must be either true or false, so we must pick only one definite value for the statements $ F \implies T$ or $F \implies F$. It's important to note that in logick we are dealing with the whole statement and things go wrong when there is contradiction with truth within the statement.

Since there are many cases when $ F \implies T$, e.g. "$3$ is even implies that $2 \times 3$ is even", we can say that the outcome is not a contradiction with the premise and we are forced to conclude that it's a true statement.

And there are many cases when $ F \implies F$, e.g. "$3$ is even implies that $3 \times 3$ is even", we can say that the outcome is not a contradiction with the premise and we are forced to conclude that it's a true statement.

In natural language statements can be vague and we don't force logical robustness upon them:

"The logical statement $p \implies q$ is nothing else than $\lnot p \lor q$" $\space \space \space$ - Hermann Weyl

Answer 12

its the the table for logical implication... To understand why this table is the way it is, consider the following example:

P-"If you get an A", Q-"then I'll give you a dollar."

The statement will be true if I keep my promise and false if I don't.

Suppose it's true that you get an A and it's true that I give you a dollar. Since I kept my promise, the implication is {\it true}. This corresponds to the first line in the table.

Suppose it's true that you get an A but it's false that I give you a dollar. Since I didn't keep my promise, the implication is false. This corresponds to the second line in the table.

What if it's false that you get an A? Whether or not I give you a dollar, I haven't broken my promise. Thus, the implication can't be false, so (since this is a two-valued logic) it must be true. This explains the last two lines of the table.

Answer 13

Intuitive/Sample based on answer may not be precise/convinced. You still need to know the back-end logic.

Here is my answer : Formula $P\implies{Q}$ is abbreviation for :$\neg{P}\vee{Q}$.

So, take this as an example : $P\implies{\neg{P}}$. If P is false , then $\neg{P}$ is true. Hence we get $F\implies{T}$ is $T$.

Another one is the "$F\implies{F}$" is $T$. which will answer the question In classical logic, why is $(p\Rightarrow q)$ True if both $p$ and $q$ are False? ,but i can not edit .

$P\implies{P}$ will be true for the case of P is false.

So, the trueth table make sense now.

Answer 14

As Henning Makholm states in his answer, the ⇒ operator is not equivalent to the usual definition of "implies".

I will add another way of looking at it. In classical logic a statement must resolve to true or false (the truth table). But using the usual definition of implies, in a couple of cases the statement will resolve to "don't know" or "unproven". So not only are the classical logic and usual definitions not equivalent, there was never any possibility for them to match.

I describe p ⇒ q using usual definitions as, "the values of p and q are consistent with the statement that p implies q".

Answer 15

I am kinda surprised nobody mentioned this: this truth table corresponds to the weakest (i.e. carries the least amount of information, i.e. being true most of the time (think about tautologies, you can’t infer anything from them since they are true no matter what/in all interpretations)) connective that modus ponens holds true.

To be more precise, what I meant is suppose we want both $p, p \rightarrow q \vdash q$ (modus ponens) and $\Gamma \vdash p \Rightarrow \Gamma \models p$ (soundness of the proof system), then we must have the row "T F F". If we additionally want implication to be the weakest¹ connective, then all other rows will have the truth value of T which gives exactly the usual truth table.

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¹ a theory $\Gamma_1$ is weaker than another theory $\Gamma_2$ iff the set(class) of models of $\Gamma_1$ is a proper superset(class) of $\Gamma_2$. If the proof system $\vdash$ is complete then this implies the deductive closure $\mathrm{Th}(\Gamma_1) \subset \mathrm{Th}(\Gamma_2)$.

Answer 16

@user701510 Conditional ($\Rightarrow$) also known as "material implication", "material consequence", or simply "implication" follows the condition 'if...then'

| p | q | p -> q |
| T | T |   T    |
| T | F |   F    |
| F | T |   T    |
| F | F |   T    |

$p \Rightarrow q$ in the best and most simple way I understand it is by giving a situation. For example, in checking a test paper.

First row implies that "If the given statement or question is right, and you gave the right answer, then you are correct."

Second row: "If the given statement or question is right, but you gave the wrong answer, then you are definitely incorrect."

Third row: "If the given statement or question is wrong (e.g. 'partially and grammatically incorrect by its sense'), but you gave the right answer (e.g. 'you get the point', 'you comprehend on the thought of what is being asked'), then you are correct."

Fourth row: "If the given statement or question is wrong (completely wrong), whatever your answer might fail, then it may be a bonus point.

I asked my professor in Discrete Structures (Mathematics), I just applied on the given condition.

Answer 17

p->q If I have chocolate, then I am happy.

T T -> T If I have chocolate, then I am happy. As initially stated.

T F -> F If I have chocolate, then I cannot be not happy, by the initial statement.
..............That's why this is false.

F T -> T If I do not have chocolate, I could still be happy
.............. (perhaps because I have a cookie).
.............. This is the one you asked about. Nobody said that p is a
.............. NECESSARY condition for q, just that it is a SUFFICIENT condition.

F F -> T If I do not have chocolate, then I could also be not happy
..............(because nothing else is making me happy).