I am studying entailment in classical first-order logic.
The Truth Table we have been presented with for the statement $(p \Rightarrow q)\;$ (a.k.a. '$p$ implies $q$') is: $$\begin{array}{|c|c|c|} \hline p&q&p\Rightarrow q\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T\\\hline \end{array}$$
I 'get' lines 1, 2, and 3, but I do not understand line 4.
Why is the statement $(p \Rightarrow q)$ True if both p and q are False?
We have also been told that $(p \Rightarrow q)$ is logically equivalent to $(~p || q)$ (that is $\lnot p \lor q$).
Stemming from my lack of understanding of line 4 of the Truth Table, I do not understand why this equivalence is accurate.
$\endgroup$ 12Administrative note. You may experience being directed here even though your question was actually about line 3 of the truth table instead. In that case, see the companion question In classical logic, why is $(p\Rightarrow q)$ True if $p$ is False and $q$ is True? And even if your original worry was about line 4, it might be useful to skim the other question anyway; many of the answers to either question attempt to explain both lines.
23 Answers
$\begingroup$Here is an example. Mathematicians claim that this is true:
If $x$ is a rational number, then $x^2$ is a rational number
But let's consider some cases. Let $P$ be "$x$ is a rational number". Let $Q$ be "$x^2$ is a rational number".
When $x=3/2$ we have $P, Q$ both true, and $P \rightarrow Q$ of the form $T \rightarrow T$ is also true.
When $x=\pi$ we have $P,Q$ both false, and $P \rightarrow Q$ of the form $F \rightarrow F$ is true.
When $x=\sqrt{2}$ we have $P$ false and $Q$ true, so $P \rightarrow Q$ of the form $F \rightarrow T$ is again true.
But the assertion in bold I made above means that we never ever get the case $T \rightarrow F$, no matter what number we put in for $x$.
$\endgroup$ 3 $\begingroup$Here are two explanations from the books on my shelf followed by my attempt. The first one is probably the easiest justification to agree with. The second one provides a different way to think about it.
From Robert Stoll's “Set Theory and Logic” page 165:
To understand the 4th line, consider the statement $(P \land Q) \to P$. We expect this to be true regardless of the choice of $P$ and $Q$. But if $P$ and $Q$ are both false, then $P \land Q$ is false, and we are led to the conclusion that if both antecedent and consequent are false, a conditional is true.
From Herbert Enderton's “A Mathematical Introduction to Logic” page 21:
For example, we might translate the English sentence, ”If you're telling the truth then I'm a monkey's uncle,” by the formula $(V \to M)$. We assign this formula the value $T$ whenever you are fibbing. In assigning the value $T$, we are certainly not assigning any causal connection between your veracity and any simian features of my nephews or nieces. The sentence in question is a conditional statement. It makes an assertion about my relatives provided a certain condition — that you are telling the truth — is met. Whenever that condition fails, the statement is vacuously true.
Very roughly, we can think of a conditional formula $(p \to q)$ as expressing a promise that if a certain condition is met (viz., that $p$ is true), then $q$ is true. If the condition $p$ turns out not to be met, then the promise stands unbroken, regardless of $q$.
That's why it's said to be “vacuously true”. That $(p \to q)$ is True when both $p$, $q$ are False is different from saying the conclusion $q$ is True (which would be a contradiction). Rather, this is more like saying “we cannot show $p \to q)$ to be false here” and Not False is True.
$\endgroup$ 6 $\begingroup$Here's a slightly different answer from the ones given.
The last line of the truth table is indeed counter-intuitive and this is exploited by the Wason Selection task. In the test, subjects are asked to solve this following puzzle:
There are 4 cards placed on a table. One side of the card has a number, while the opposite side is only coloured. The visible faces of the cards show 3, 8, red and blue. Which card(s) should you turn over in order to test the truth of the proposition that if a card shows an even number on one face, then its opposite face is red?
In Wason's original experiment, only 10 percent responded correctly. Most failed to list the blue card as the card that also must be turned over (apart from the card with number 8). Now, suppose we turn the blue card. Only if the other side fails to have an even number, would the proposition be true. Why? Because if the other side was even, then you'd have a card with even number on one face whose opposite face was not red. You can look at this as the intuition behind the last line of the truth-table. To correctly test the verity of the proposition, we must check that when the consequent is False, then the antecedent must be False too.
$\endgroup$ 1 $\begingroup$The usual truth table interpretation of $\implies$ certainly does not capture the connotations of the English word "implies." This has sparked a number of attempts to define a notion of "strong implication" that is more faithful to the informal meaning, and in particular is it is one motivation behind modal logic.
A partial answer to the OP's question goes as follows. Suppose that we decide that the truth value of $A \implies B$ is to be completely determined by the truth values of $A$ and $B$ (in jargon, that $\implies$ is to be truth-functional.) That already does violence to the ordinary language meaning of "implies," but let us go on.
What value shall we assign to $A \implies B$ when $A$ is false and $B$ is true? What value shall we assign when $A$ is false and $B$ is false? A brief examination of the alternatives, aided by the responses already posted, shows that any alternative to the standard truth value assignment would be worse.
$\endgroup$ $\begingroup$One reason: $p$ implies $q$ should be equivalent to its contrapositive, not $q$ implies not $p$.
$\endgroup$ 1 $\begingroup$I'm surprised nobody has given this response yet. It is my current favorite. (paraphrased from: )
I'd say "implies" means the same as "subset" in set theory. That is, when you say
If it rains, then the ground gets wet
you mean
The set of times when it rains is a subset of the set of times when the ground gets wet.
So, since the empty set is a subset of any set, a false statement implies any statement.
Of course, a perfectly valid (and annoyingly common) answer is "That's just how we defined it, and we can define things however we want", but hopefully the above explanation addresses the heart of the matter better.
$\endgroup$ 3 $\begingroup$The short answer is: 'This is the definition, live with it!' While that may sound haughty, it emphasizes the principle that in math we occasionally need to define a precise meaning to a word, and natural languages may fall short in providing a suitable one, so it is up to us the users of math to absorb a possibly new meaning. It is the price we need to pay so that a statement will have a precise meaning. The longer answer is that this meaning is more useful in writing mathematical results. Let me elaborate a bit.
Yet another way of looking at it is the following. It emerges, when we add one (or more) variable(s) $x$ ($y,z,\ldots$) to the language (really moving to predicate logic, but that's the language mathematical truths are written in). So instead of a proposition $p$ with a definite truth value we have a statement $p(x)$ whose truth depends on the value assigned to the variable $x$. We really want the implication
$$p(x)\Rightarrow q(x)$$
at the level of predicates to mean $\forall a: (p(a)\rightarrow q(a))$, where $a$ ranges over the elements of whatever set is relevant in the context. This is what is needed to express the usual mathematical results. In natural language $p(x)\Rightarrow q(x)$ should have either of the following equivalent meanings: 'if $p(x)$ is true, then so is $q(x)$', or '$q(x)$ is true whenever $p(x)$ is'. Notice that when a mathematician claims '$p(x)\Rightarrow q(x)$' she/he is not claiming anything about the truth of $q(a)$ unless $p(a)$ holds. So for example the statement $x>0\Rightarrow 2x>0$ as well as the statement $x>0\Rightarrow x+1>0$ are both valid implications, when the context tells us that $x$ is a real number, right? In both implications $p(x)$ means $x>0$. In the first case $q(x)$ means $2x>0$ and in the latter example $q(x)$ means $x+1>0$. Therefore the former reads in natural language: $2x$ is positive, whenever $x$ is, and the latter reads: if $x$ is positive, then so is $x+1$.
The first of these implications forces us to define the fourth line of the truth table the way it is done. For otherwise the implication would break down, because $p(-1)\rightarrow q(-1)$ would then be false, as both $p(-1)$ and $q(-1)$ are false. The latter implication (that we also want to be true) forces us to define the third line the way it is done, because $p(-1/2)$ is false but $q(-1/2)$ is true.
My point here is that the need to define it this way is clearer at the level of implications between predicates. At the level of propositions it is mostly a definition, but at the level of predicates, we are really making deductions. My education in formal logic is somewhat lacking, so please comment on my errorneous use of terms, and I will edit. I am approaching this question as a teacher of freshman calculus/algebra :-)
$\endgroup$ $\begingroup$This is really just a convention, as everyone else has noted. The way to remember it is with a negative definition of implication: $P \to Q$ is the proposition that is true unless $P$ is true and $Q$ is false.
I think this is a nice example: "if I pass my exams, then I will get drunk". If you passed your exams and then ended up in the gutter with a bottle of Tequila in hand, you spoke truly. If you passed your exams then went home and had an early night, shame on you, your statement was false. If you failed your exams, then we'll never know what you would have done if you'd passed them, but it hardly seems fair to call you a liar.
$\endgroup$ 1 $\begingroup$One can use intuitive reasoning to figure out that it is best if 'False implies False' is 'True'.
Consider the informal causative meaning of 'implies', $P \rightarrow Q$ means that $P$ causes $Q$. ANother way to say this is that the appearance of $P$ causes $Q$ to appear.
Now, to judge whether $P \rightarrow Q$ is a correct statement, you need to check all the possibilities of when $P$ and $Q$ appear or not (this is the truth table).
For the case of interest, $P$ does not appear and also $Q$ does not appear, is this case acceptable according to $P \rightarrow Q$? Yes, it is, because the statement '$P$ causes $Q$' is perfectly OK if neither of them appear. $Q$ can be caused by many things, so can $\neg Q$; if $P$ is not the case, then it doesn't say anything about $Q$. So, $P \rightarrow Q$ is acceptable even if neither are true.
The usual difficulty with $\rightarrow$ is the case $P=$ False and $Q =$ True, because the English (and most natural languages) 'if-then' has the tendency to be interpreted as 'if-and-only-if' the value of $P$ is the same as the value of $Q$.
To connect $P\rightarrow Q$ with $\neg P \lor Q$, the above explanation says that when $Q$ is true, the value of $P\rightarrow Q$ is the same as the value of $P$. But if $P$ is false, anything goes for $Q$, so $F \rightarrow Q$ is True. And that's the same for $\neg P \lor Q$.
$\endgroup$ $\begingroup$To understand consider this question: How must the world be designed to make this statement ( $p \implies q$ ) true?
There are two options:
- Option one: $p$ is not true. Then our statement is always true, because we do not impose anything, as implication means "if p is true then...." (explains lines 3 and 4)
- Option two: if $p$ is true, $q$ must be true too to fulfill the statement (lines 1 and 2)
Combining these we get $\sim p \vee q$.
$\endgroup$ $\begingroup$An implication holds if "it is impossible for the premises to be true but the conclusion false." Given that your premise is known to be false, it is impossible for (the premise to be true) AND (the conclusion false) because the premise isn't true(so it fails the first condition).
Similarly, if $q$ is true then $p \implies q$ must be true: it is impossible for (the premise to be true) AND (the conclusion false) because the conclusion isn't false(so it fails the second condition).
$\endgroup$ $\begingroup$Why is the statement (p -> q) True if both p and q are False?
Since one of the main axioms in mathematics is Modus Ponens which allows one to infer truths from an implication the idea was to define implication in such a way that all false statements are rendered useless. Obviously the easiest and most natural way to do this is to make them able to imply anything, right or wrong. This is known as "ex falso sequitur quodlibet" i.e. false statements imply anything (and are thus useless).
Stemming from my lack of understanding of line 4 of the Truth Table, I do not understand why this equivalence is accurate.
I assume you know the truth tables for logical negation $\neg$ and logical disjunction $\vee$. From there it's a very easy exercise to show that for any two statements $p,q$ we have $(p\implies q)\iff(\neg p\vee q)$.
$\endgroup$ 10 $\begingroup$If the fourth line were F, then proving p implies q would prove q. I'm not sure why the third line is any less troubling.
$\endgroup$ 1 $\begingroup$$p$ = you eat your meat
$q$ = you can have pudding
You don't eat your meat. ($p$ is false)
You can't have any pudding. ($q$ is false)
Your question is why, under these circumstances, "if you eat your meat, you can have pudding" is true. Sometimes just substituting in the right $p$ and $q$ makes the answer obvious.
$\endgroup$ 1 $\begingroup$If we consider True to be "more true" than False, (p⇒q) expresses the statement that q is at least as true as p, or q is not less true than p. This is a fundamental requirement for valid reasoning, since the most important reason we use logic at all is to keep us from somehow introducing a false conclusion when we have started with true assumptions. P may represent a whole list of premises all assumed to be true.
Considering the four possibilities:
1) When p is true and q is true, q is at least as true. (p⇒q) checks as true, meaning that it's a valid statement because we haven't introduced a false conclusion starting with true premises.
2) When p is true and q is false, q is NOT at least as true as p and IS less true. Then, and only then, we can say that (p⇒q) is false, meaning that it's not valid because we have introduced a false conclusion starting with true premise.
3) When p is false and q is true, q is at least as true as p. It's more true, but that's irrelevant. (p⇒q) then checks as True, meaning that it's a valid statement because we haven't introduced a false conclusion starting with true premises. It doesn't matter that any of the premises were false to begin with.
4) Likewise, when p is false and q is false, (p⇒q) then checks as true, meaning that it's a valid statement. q is at still at least as true as p; it's not less true. We still haven't introduced a false conclusion starting with true premises.
(P =>Q) may be considered true when all we know is that P is false or Q is true. But in that case, it's trivially true and useless. We can't use the fact to conclude anything else.
(P =>Q) may used be to express the stated relationship as a fact: For some reason, we know that q is at least as true as p. This may be only because we have assumed or declared it and wish to explore the consequences of that assumption.
(P =>Q) may be used to express the fact that we have assumed p and somehow derived q as a conclusion. But that's not the only thing it can mean, so many logicians prefer to use other symbols and other terminology for that particular meaning.
$\endgroup$ $\begingroup$I always remember it this way:
If you have false facts and want to make a conclusion from those, your conclusion might be true or false ($true \lor false \equiv true$).
$\endgroup$ $\begingroup$If "men live in the moon" then they have four eyes. Is an implication of two wrong (propositions) premises that is perfectly true.
$\endgroup$ 1 $\begingroup$Theorem
For any true-or-false propositions $A$ and $B$, we have:
$\neg A \implies [A\implies B]$
Proof
Suppose $\neg A$
Suppose $A$
Suppose (to the contrary) $\neg B$
Obtain the contradiction $A \land \neg A$ from (2) and (1).
Conclude by contradiction that $\neg\neg B$ from (3) and (4).
Remove $\neg\neg$ to obtain $B$ from (5).
Conclude that $A\implies B$ from (2) and (6).
Conclude as required that $\neg A \implies [A\implies B]$ from (1) and (7).
Comment
This result would seem to be the inevitable consequence of allowing the Law of the Excluded Middle (line 6) and proofs by contradiction (line 5), which, I think, we do even in natural language. In summary, all things follow from a falsehood.
EDIT: More on this topic in the posting "If Pigs Could Fly" at my math blog.
$\endgroup$ 14 $\begingroup$First, forget about the words "if" and "then", or even "implies". No, really.
Instead I'm going to start with something that looks completely unrelated, namely infinite sums. Don't worry, eventually it's going to lead to a nice explanation for why $\to$ has the truth table it does, but there's some ground to cover first to give the right perspective.
Infinite sums.Suppose we want to write down the sum of reciprocals of squares of all the primes:$$ \frac1{2^2} + \frac1{3^2} + \frac1{5^2} + \frac1{7^2} + \frac1{11^2} + \cdots $$(This sum converges to a number near 0.452247 that doesn't seem to have a nice closed expression, but the precise value will not concern us here; just how to write down the summation).
In order to make the $\cdots$ precise, we'd want to write it with a $\sum$ sign somehow. What one writes in practice is something ad-hoc like$$\sum\limits_{p\text{ prime}} \frac1{p^2} \tag{1}$$but suppose the only summation operator we had was $\sum\limits_{n=1}^{\infty}$. We definitely can't write$\sum\limits_{n=1}^{\infty} \frac1{n^2}$because that would mean$$ \frac1{1^2} + \frac1{2^2} + \frac1{3^2} + \frac1{4^2} + \frac1{5^2} + \frac1{6^2} + \frac1{7^2} + \frac1{8^2} + \frac1{9^2} + \cdots $$which has terms we don't want. We need a way to "cross out" the terms that come from a non-prime $n$:$$ \require{cancel} \cancel{\frac1{1^2}} + \frac1{2^2} + \frac1{3^2} + \cancel{\frac1{4^2}} + \frac1{5^2} + \cancel{\frac1{6^2}} + \frac1{7^2} + \cancel{\frac1{8^2}} + \cancel{\frac1{9^2}} + \cdots $$There's a straightforward way to "cross out" a term in a sum: just replace it by zero, giving$$ 0 + \frac1{2^2} + \frac1{3^2} + 0 + \frac1{5^2} + 0 + \frac1{7^2} + 0 + 0 + \cdots $$Thus we can write our sum as$$ \sum_{n=1}^\infty \begin{cases} 1/n^2 & \text{when $n$ is prime} \\ 0 & \text{otherwise} \end{cases} \tag{2} $$This works because $0$ is a netural element for the $+$ operation.
What we'd really like is to have a nice compact notation for the case analysis in the general term of (2). One possibility is the Iverson bracket which would let us write$$ \sum_{n=1}^\infty [n\text{ is prime}]\cdot\frac{1}{n^2} $$but unfortunately that notation hasn't really caught on, so in practice we need to write things like (1) if we want to be understood.
Infinite disjunction.Moving on from analysis to logic, suppose we want to express the claim: There exists an even prime. Intuitively this is an infinite disjunction,$$ \text{even}(2) \lor \text{even}(3) \lor \text{even}(5) \lor \text{even}(7) \lor \cdots $$
Fortunately logic provides us with a way of writing infinite conjunctions: $\exists$ is to $\lor$ like $\sum$ is to $+$! But we run into a similar problem as before, because if we just write$$ \exists n\in\mathbb N: \text{even}(n) $$we would get$$ \text{even}(1) \lor \text{even}(2) \lor \text{even}(3) \lor \text{even}(4) \lor \text{even}(5) \lor \text{even}(6) \lor \cdots $$which is not what we wanted to say (even though both claims happen to be true).
The solution to this is also the same as before: we cross out the disjuncts we're not interested in:$$ \cancel{\text{even}(1)} \lor \text{even}(2) \lor \text{even}(3) \lor \cancel{\text{even}(4)} \lor \text{even}(5) \lor \cancel{\text{even}(6)} \lor \cdots $$
With $\lor$, the way to strike out a disjunct is to replace it by $\mathsf{false}$, because $\mathsf{false}$ is to $\lor$ like $0$ is to $+$. So we get$$ \mathsf{false} \lor \text{even}(2) \lor \text{even}(3) \lor \mathsf{false} \lor \text{even}(5) \lor \mathsf{false} \lor \cdots $$Or, going back to quantifiers:$$ \exists n\in\mathbb N: \begin{cases} \text{even}(n) & \text{when $n$ is prime} \\ \mathsf{false} & \text{otherwise} \end{cases} \tag{3} $$
This is still a bit more cumbersome to write than we'd really like. Ideally we ought to have a logical connective to write the case analysis in (3) compactly, something like the Iverson bracket but for logic:$$ \text{crossout}(P(x),Q(x)) \equiv \begin{cases} P(x) & \text{when }Q(x) \\ \mathsf{false} & \text{otherwise} \end{cases} \tag{4} $$Surprise! If we write out the truth table for the case analysis in (4) what we get is exactly the same as the truth table for $Q(x)\land P(x)$. So we can finally rewrite our claim on a single line as$$ \exists n\in\mathbb N: n\text{ is prime} \land \text{even}(n) \tag{5} $$
Infinite conjunction.We're getting closer. Suppose we want to write down Fermat's little theorem: If $p$ is a prime, then $p$ divides $a^p - a$.
This is morally an infinite conjunction:$$ (2 \mid a^2-a) \land (3 \mid a^3-a) \land (5 \mid a^5-a) \land (7 \mid a^7-a) \land \cdots $$
Now $\forall$ is to $\land$ like $\exists$ is to $\lor$ (and like $\sum$ is to $+$), so we can repeat the procedure from the previous section, where we cross out non-prime conjuncts and end up with$$ \forall n\in\mathbb N: \begin{cases} 2\mid a^n-a & \text{when $n$ is prime} \\ \mathsf{true} & \text{otherwise} \end{cases} \tag{6} $$Why is there now a $\mathsf{true}$ in the "otherwise" line instead of a $\mathsf{false}$? That's because the neutral element for $\land$ is $\mathsf{true}$ rather than $\mathsf{false}$.
So we can't reuse the final step that took us to (5) above; we need a new connective$$ \text{crossout}'(P(x),Q(x)) = \begin{cases} P(x) & \text{when }Q(x) \\ \mathsf{true} & \text{otherwise} \end{cases} $$Perhaps that would be $\lor$, since $\land$ worked the last time? Let's work out its truth table:$$\begin{array}{|c|c|c|} \hline p&q&\text{crossout}'(p,q)\\ \hline \mathsf{true}&\mathsf{true}&\mathsf{true}\\ \mathsf{true}&\mathsf{false}&\mathsf{false}\\ \mathsf{false}&\mathsf{true}&\mathsf{true}\\ \mathsf{false}&\mathsf{false}&\mathsf{true}\\\hline \end{array}$$Even more surprise! This is not the truth table of $p\lor q$, but it is the truth table of $p\to q$.
Finally $\to$ appears, with an explanation of where it comes from. It is simply the symbol we have decided to use for $\text{crossout}'$, no more and no less, such that finally write our claim as
$$ \forall n\in\mathbb N: n\text{ is prime} \to (n\mid a^n-n) \tag{7} $$
This finally gives us an answer to the question: $\mathsf{false} \to \mathit{whatever}$ must be $\mathsf{true}$ because $\mathsf{true}$ is how we cross out an operand to $\land$!
The word "implication" is all that remains to be justified, and the story for that is honestly a bit weaker than the story about why the truth table is what it is. The best I can offer is that (7) expresses the claim$$ \text{if $p$ is prime then $p\mid a^p-a$} $$using a $\to$ and a $\forall$ -- but both of these symbols are really involved in expressing the "if...then" meaning. They work as a team, not as unrelated players.
We could avoid having a $\to$ connective with a "weird" truth table at all, if instead of $\forall$ we had a combined construction$$ \bigwedge_{p\in\mathbb N,~ p \text{ prime}} (p \mid a^p-p) $$reminiscent of (1). But that's not how the convention in symbolic logic has ended up.
Duality.The moral of the above story is that $\to$ is made to work together with $\forall$. It will yield very weird meanings if used in the combination $\exists x(\cdots \to \cdots)$. And indeed $\to$ is essentially never seen in that context in mathematics, other than in parlor tricks such as the Drinker's Paradox where the weirdness is exactly the point.
$$ \to \text{ goes with } \forall \\ \text{where} \\ \land \text{ goes with } \exists $$
How come the existing $\land$ worked to cross out disjuncts for $\exists$ but we needed to invent a new $\to$ symbol to cross out conjuncts for $\forall$?
One way to understand it is to look at the shape De Morgan duality takes in this perspective:$$ \begin{align} \neg\exists x: Q(x) \land P(x) &~\equiv~ \forall x: Q(x) \to \neg P(x) \\ \neg\forall x: Q(x) \to P(x) &~\equiv~ \exists x: Q(x) \land \neg P(x) \end{align}$$If we had just applied the ordinary De Morgan's laws blindly, we'd instead get$$ \neg\exists x: Q(x) \land P(x) ~\equiv~ \forall x: \neg Q(x) \lor \neg P(x) $$which is certainly valid too, as far as it goes.
However, here we're viewing $Q(x)$ as a "guard" formula that tells us which values of $x$ we're interested in at all. It feels backwards that the negation of the formula would include speaking about $x$ that we're not interested in, so instead we'll often prefer a duality rule that lets $Q$ remain non-negated to the right of the $\equiv$. The values for $x$ we're interested in are still the same; we're just asking a different question about them. And that is exactly what $\forall x:Q(x)\to\neg P(x)$ achieves for us.
$\endgroup$ 1 $\begingroup$My intuitive way of understanding it goes as follows: is the situation in which both P and Q are false, compatible with the statement that P implies Q?
Or else: in a universe in which P implies Q, would it be possible to observe both not P and not Q?
This explanation has not been mentioned yet:
If you want to use modus tollens then you need (p⇒q) to be true if $p$ and $q$ are false, otherwise modus tollens cannot be defined.
$\endgroup$ $\begingroup$Let's consider the following statement:
If it rains I stay at home
Now consider this:
It rains and I stay at home - This does not contradict my statement and corresponds to T implies T equals T in the truth table.
It rains and I go for a walk - This does contradict my statement and corresponds to T implies F equals F in the truth table.
It does not rain but I still decide to stay at home - No contradiction. F implies T equals T.
It does not rain and I go to movies - No contradiction. F implies F equals F.
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 3 (and also 4) shows that we should therefore have that $F \to F = T$
Also note that line 1 forces $T \to T = T$, and that line 2 forces $F \to T=T$, 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.
$\endgroup$ 1
