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.

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.

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    $\begingroup$ I assume for now that when you say "p is not a condition for q", you mean p is not a necessary condition for q. And so it is not. "If p then q" does not mean p is necessary for q; it means p is sufficient for q. $\endgroup$ Commented Oct 7, 2011 at 23:00
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    $\begingroup$ Consider the following (totally informal) proposition: if it rains tomorrow, I'll bring my umbrella. However, the fact that I'll bring my umbrella doesn't imply it will rain tomorrow. $\endgroup$ Commented Oct 7, 2011 at 23:09
  • $\begingroup$ If pigs could fly... $\endgroup$
    – ahorn
    Commented Sep 13, 2015 at 17:24
  • $\begingroup$ Is there any relation with the principle of explosion and the last two values of the truth table? It seems a very weird coincidence that you can conclude both statements as if the principle of explosion was used to justify the last two entries in the truth table. $\endgroup$ Commented Jul 5, 2017 at 14:42
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    $\begingroup$ It might help to broadly conceptualize the notion of implication $\implies$ as meaning "we have no evidence against" rather than "we have positive evidence for". (This is just a heuristic though - you should of course consult the actual definition of $\implies$ when you're reading or writing formal math and you aren't sure how to interpret an expression or whether the use of that symbol is legitimate.) $\endgroup$
    – tparker
    Commented Jan 18, 2018 at 6:10

18 Answers 18


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".

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    $\begingroup$ It does not violate the principle but we cannot be sure that the soda machine works so we cannot be sure that the principle is true. However, it seems that I have misinterpreted the purpose of the truth table. $\endgroup$
    – user701510
    Commented Oct 8, 2011 at 5:05
  • 17
    $\begingroup$ @user701510, the truth table is designed to be useful under a quantifier, so that we can define that "the soda machine works" means "every time someone walks past the machine, it is true that if they put money in, a soda comes out". We want to interpret the if-then such that we don't consider the machine to not work simply because some people walk past without buying from it. $\endgroup$ Commented Oct 8, 2011 at 17:09
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    $\begingroup$ Your soda example has made sure that I never forget the truth table of p->q! Thanks $\endgroup$ Commented Mar 11, 2015 at 0:54
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    $\begingroup$ maybe this is a similar complaint as user701510 but just because in everyday logic I would probably think that I don't know if the implication is true or false. I would just call it unknown. My confusion is why did we not choose such an interpretation for the last two entries as it seems more in line with everyday common sense. Is very often that formalism in mathematics just try to make some intuitive concept/idea formal. Why would this one be different? $\endgroup$ Commented Jul 5, 2017 at 14:41
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    $\begingroup$ I actually could and would argue that machine is broken if it gave me a bottle but I didn't put money. Since it should only give soda if and only if I put in money. Oh. But if I state that the machine should only give soda when money is given, then that is not implication but biconditional right? $\endgroup$
    – user500668
    Commented May 2, 2018 at 16:53

$p\Rightarrow q$ 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.

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    $\begingroup$ +1 Very nice! Exactly the problem I had for years. People try to define it with "If p, then q" but that is NOT the definition :-). $\endgroup$
    – TFuto
    Commented Mar 28, 2014 at 17:55
  • 2
    $\begingroup$ I know that you say that the last two entries are true by definition (and your intuitive justification is something I am familiar with). However, what I've always been curious to know is, what would be the consequences in mathematics if we actually defined them false. My intuitive justification would be that starting from a false premise is non-sense and you shouldn't have a true implication out it. Regardless if thats good, I always thought the principle of explosion was the reason we defined the table that way, since otherwise we would abandon even more fundamental building blocks of logic. $\endgroup$ Commented Jul 6, 2017 at 14:24
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    $\begingroup$ @Pinocchio: The basic motivation is that we want to be able express, for example, "all dogs are mammals" as $\forall x(\mathit{dog}(x)\Rightarrow\mathit{mammal}(x))$ -- where $\forall x$ has the easily explainable semantics of claiming something about everything in existence. If those lines did not produce true, $\forall x(\mathit{dog}(x)\Rightarrow\mathit{mammal}(x))$ would become false in a world that contains cats (not dogs, yet mammals) or fish (neither dogs nor mammals). $\endgroup$ Commented Jul 6, 2017 at 15:16
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    $\begingroup$ Also, if the result column was "TFFF", $\Rightarrow$ would have exactly the same truth table as $\land$, in which case it would be silly to have an additional symbol for the same thing. If the result column was "TFFT", $\Rightarrow$ would mean the same as $\Leftrightarrow$. And if it was "TFTF", $p\Rightarrow q$ would mean the same as $p$ itself -- which would be even more pointless. $\endgroup$ Commented Jul 6, 2017 at 15:18
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    $\begingroup$ @Pinocchio: You misunderstood my first point. It is not about "if there are no dogs" -- it is if there is anything that is not a dog, even if there are also dogs. Suppose the universe is $\{A,B,C\}$ where $A$ and $B$ are dogs and $C$ is a fish. Then, under your proposed truth table, ${\it dog}(C)\Rightarrow{\it mammal}(C)$, which is $F\Rightarrow F$, would be $F$, and therefore $\forall x({\it dog}(x)\Rightarrow{\it mammal}(x))$ would also be false. It wouldn't help that ${\it dog}(x)\Rightarrow{\it mammal}(x)$ is true for $x=A$ and $x=B$, because $\forall x$ demands truth for everything. $\endgroup$ Commented Jul 6, 2017 at 18:40

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.

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) $T$

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.


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

  • 3
    $\begingroup$ Your link is not working any more! :) $\endgroup$ Commented Sep 1, 2016 at 12:07

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.

  • 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

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

  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.


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.


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.


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.


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

  • $\begingroup$ Am I right that "outcome" refers to q and "premise" refers to p? What if I modify your first example as "3 is even implies 3 is odd". This is a "F => T" case but "3 is odd" (i.e., outcome) contradicts "3 is even" (i.e., premise). What should be the truth value of the whole statement? (I'm still confused but your answer already sheds some light, especially the quote from Hermann. Thanks!) $\endgroup$
    – yaobin
    Commented Jul 22, 2023 at 22:23

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).


@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.

  • $\begingroup$ Welcome to math.se. Tip: you can write click on equations you see in other posts, choose "show math as", "tex commands", to see how to input equations. Enjoy~ $\endgroup$
    – DanielV
    Commented Jul 9, 2014 at 6:43

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.


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".


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.

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 $\mathrm{Th}(\Gamma_1) \subset \mathrm{Th}(\Gamma_2)$.

  • $\begingroup$ The term to search for here is proof-theoretic harmony and proof-theoretic semantics. $\endgroup$
    – Poscat
    Commented Mar 13 at 12:03

Consider the implication, "If it is raining ($R$), then it is cloudy ($C$)." Symbolically, we can write:

$~~~~~~R \implies C$

There seems to be some confusion about what this means in classical (propositional) logic. It does not mean that rain causes cloudiness. It also does not mean that it is always cloudy when it is raining. It simply rules out the possibility that it is currently both raining and not cloudy.

$~~~~~~R \implies C~~\equiv \neg (R \land \neg C)$

Here is the truth table: **

enter image description here

This is not about causality or the historical record. It is about a single instant in time.

Addressing the original question here, if it is not raining ($\neg R$), then the implication ($R\implies C$) will be true regardless of whether it is cloudy or not.

This form of argument is rarely, if ever, used in daily discourse since we rarely consider the implications of propositions known to be false. It is, however, routinely used in very technical arguments, e.g. in mathematical proofs.

The Principle of Vacuous Truth: Generally speaking, if the antecedent of an implication is false, then that implication will be true. We cannot, however, infer from this that the consequent is true.

** Truth table in plain text:

R C R=>C






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