# Assumed True until proven False. The Curious Case of the Vacuous Truth

Given two statements, $P$ and $Q$, and the logical connective, $\implies$, the truth table for $P \implies Q$ is:

$$\begin{array}{ c | c || c | } P & Q & P\Rightarrow Q \\ \hline \text T & \text T & \text T \\ \text T & \text F & \text F \\ \text F & \text T & \text T \\ \text F & \text F & \text T \end{array}$$

Lines one and two are quite clear. What is ambiguous is lines three and four.

One explanation as to why $P \implies Q$ is true when $P$ is false, provided by Velleman goes:

Let $P(x)$ be the statement $x>2$ and $Q(x)$, $x^2 > 4$. When $x=3$, $P$ is true, and $Q(x) = 9$ thus $Q$ is true. If $P(x) = 1$ then $Q(x) = 1$ and they are both false. If $P(x) = -3$ then $Q(x) = 9$ and so the statement is true.

This explanation was quite unsatisfactory to me. Looking at Enderton, we have:

For example, we might translate the English sentence, ”If you're telling the truth then I'm a monkey's uncle,” by the formula ($V \implies M$). 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 ($p\implies q$) 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 $q$.

Though a significant improvement over the Velleman explanation, I still feel uncomfortable with it.

Really, it seems we can conjure up as many silly counter-examples as we like, such as:

If pigs can fly, then I can walk on water.

Though following the above truth table, the implication is that my ability to walk on water is true.

After considering it, it seems to me that $\implies$ is only meaningful when $P$ is true, then we can look at it in relation to $Q$. However, if $P$ is false, then we actually know nothing about the relationship between $P$ and $Q$. This would give a truth table of:

$$\begin{array}{ c | c || c | } P & Q & P\Rightarrow Q \\ \hline \text T & \text T & \text T \\ \text T & \text F & \text F \\ \text F & \text T & \text ? \\ \text F & \text F & \text ? \end{array}$$

where the $?$ denotes that given $P$ we actually don't know anything about $\implies$.

Thus. one way to clear this up would be to assume that $? = T$. Thus the vacuous truth is a "definition of convenience" in a sense.

The above is my take on the matter.

Could someone provide some clarification on the conditional logical connective?

• $P \Rightarrow Q$ is just $Q \lor \lnot P$. There's nothing causal in it. So if the premise $P$ is false, $\lnot P$ is true, hence $\lnot P \lor Q$. – Daniel Fischer Jul 9 '13 at 21:36
• Pigs can fly (just drop it from shed's roof). In a more serious tone: that is the precise meaning of implication, it does nothing when the premise is false. There are many connectives in math, but always using a two-way conjunction would be cumbersome. It is frequent case that formula works only if some conditions are met, that is, if that conditions are not met, the formula is meaningless. For example, "$ax=bx$ might be reduced to $a=b$" works only if $x \neq 0$, however, if $x = 0$ the reduction is false, but it still might be true that $a=b$. Hence, $x \neq 0$ implies that reduction. – dtldarek Jul 9 '13 at 21:53
• $\rm Then$ $\neq$ $\rm Than$. – Pedro Tamaroff Jul 9 '13 at 22:10
• "If pigs can fly, than I can walk on water. Though following the above truth table, the implication is that my ability to walk on water is true." Not at all. It is true that if pigs can fly, than you can walk on water. It is false that pigs can fly. This says nothing about whether it is true or false that you can walk on water. – Rahul Jul 9 '13 at 23:07
• See math.stackexchange.com/questions/48161/… and the "linked questions" on the side of that page. Note that lines 3 and 4 in the truth table for $p\implies q$ are not at all ambiguous. The real questions here are: $(1)$ why do we choose to have a logical connective with the properties this one has, $(2)$ why do we choose to call it "implies" and $(3)$ what relation does it have with the common-sense interpretation of the word "implies" in the English language? – blue Jul 9 '13 at 23:17

One can, correctly, assign the truth-value of true to the statement $P\implies Q$ whenever $P$ is false, or whenever $Q$ is true. $P\implies Q\,$ is false if and only if both $P$ is true and $Q$ is false. That covers all the cases. So we can say that $P\implies Q$ is true, unless "proven false", by which I mean to say:

$P \implies Q$ is true if and only if it is not the case that both $P$ is true and $Q$ is false.

What we can also say is that in classical logic and in math, it is a mistake to attribute any sort of causal relationship between $P$ and $Q$ when writing or reading an implication $P\implies Q$. Put differently, $P\implies Q$, by itself, does not imply any causal relationship between $P$ and $Q$: It is defined to convey nothing more, and nothing less, than is conveyed by the statement: $\;\lnot P \lor Q$, or if you prefer, it tells us nothing more (and nothing less) than what is conveyed by the statement: $\;\lnot(P \land \lnot Q)$.

Your concern is not trivial, nor are you alone in being "bothered" by that lack of some stronger relationship between $P$ and $Q$. There are logics, such as relevance logic which aim to capture aspects of implication that are ignored by the "material implication" operator in classical truth-functional logic, requiring some sort of relevance between antecedent and conditional of a true implication. See also the Wikipedia entry entitled: Paradoxes of material implication for more on "alternate" non-classical logics.

"... the vacuous truth is a "definition of convenience" in a sense."? No, not mere convenience. There are strong pressures that push towards making this choice of truth-values in lines 3 and 4 of the truth-table for the conditional. Here's one that others haven't mentioned.

One thing mathematicians need to be very clear about is the use of statements of generality and especially statements of multiple generality – you know the kind of thing, e.g. the definition of continuity that starts for any $\epsilon$ ... there is a $\delta$ ... And the quantifier-variable notation serves mathematicians brilliantly to regiment statements of multiple generality and make them utterly unambiguous and transparent.

Quantifiers matter to mathematicians, then: that's uncontentious. OK, so now think about restricted quantifiers that talk about only some of a domain (e.g. talk not about all numbers but just about all the even ones). How might we render Goldbach's Conjecture, say? As a first step, we might write

$(\forall n \in \mathbb{N})$(if $n$ is even and greater than 2, then $n$ is the sum of two primes)

We restrict the quantifier here by using a conditional. So now think about the embedded conditional here. What if $n$ is odd, so the antecedent of the conditional is false??? If we say this instance of the conditional lacks a truth-value, or may be false, then the quantification would have non-true instances and so would not be true! But of course we can't refute Goldbach's Conjecture by looking at odd numbers!! So, if the quantified conditional is indeed to come out true when Goldbach is right, then we'll have to say that the irrelevant instances of the conditional with a false antecedent come out true by default. In other words, the embedded conditional will have to be treated as a material conditional.

So: to put it a bit tendentiously and over-briefly, if mathematicians are to deal nicely with expressions of generality using the quantifier-variable notation they have come to know and love, they will have to get used to using material conditionals too.

• I don't see much relevance to the question here. The question concerned propositions and propositional logic, but you introduced quantification theory. – Doug Spoonwood Jul 10 '13 at 16:40
• The question was: is the truth table for the conditional in some sense "a definition of convenience". I was suggesting, precisely, that looking at how the conditional behaves in quantificational logic throws some light on this question. That's why it is relevant. – Peter Smith Jul 10 '13 at 19:24

You can interpret $P \Rightarrow Q$ as meaning "whenever I've written $P$ down in a proof, I can subsequently write $Q$ down in a proof." (In other words, we're defining implication as "the thing that satisfies modus ponens.") Certainly you can always write down a tautology in a proof, so $Q$ can always be $T$. Similarly, if you write down a false statement in a proof then you can prove anything, so $P$ can always be $F$. The only thing you can't do is start with true statements and end up with false statements, so $P$ can't be $T$ if $Q$ is $F$.

• "if you write down a false statement in a proof then you can prove anything" - doesn't this rely on $\implies$ having the properties that it does in lines 3 and 4 of its truth table? In other words, this answer seems to simply restate the facts that OP is asking for an explanation of. – blue Jul 9 '13 at 23:23
• Okay, I can say something different there too. It should always be the case that $P \Rightarrow P$ by the above definition. – Qiaochu Yuan Jul 9 '13 at 23:25

Your quote from Velleman, as it stands, is irremediably confused and nonsensical; so you are absolutely right in finding it "quite unsatisfactory".

In mathematics, it is possible to derive constructively any statement, true or false, from a false statement. For example, from "$2+2=5$", by subtracting $4$ from each side, we get "$0=1$", from which it is fairly straightforward to derive pretty well any statement you like. Outside mathematics, a statement such as "If pigs can fly, then Elvis Presley is president of the United States" can never be shown to be false, because to do so would require finding a flying pig, which will never happen. We can put such statements into the category "unfalsifiable" in distinction from "true", but generally nothing bad will happen if we lump these two categories together. The extent to which one is prepared to identify the unfalsifiable with the true is largely a matter of philosophical taste.

There are many good answers here, but I think it's worthwhile to address your claim directly. The fact that the implication $P\implies Q$ is true does not mean that the consequent $Q$ is true.

You state:

Really, it seems we can conjure up as many silly counter-examples as we like, such as:

If pigs can fly, than I can walk on water.

Though following the above truth table, the implication is that my ability to walk on water is true.

However, the two relevant lines in the truth table (where $P$ is false) are:

$$\begin{array}{ c | c || c | } P & Q & P\Rightarrow Q \\ \hline \text F & \text T & \text T \\ \text F & \text F & \text T \end{array}$$

A clear interpretation of this would be "if the implication is true and the antecedent is flase, then the consequent may be true or may be flase, i.e. we don't know."

If you want to try and explore non-truth functional connectives go right ahead. But first, do you really want the weak law of identity expressible as "if p, then p" to fail?

If you want ⇒ to come as truth-functional, the following makes for a different answer than the other ones. First formation rules:

1. All lower case letters of the Latin alphabet qualify as formulas.
2. If $\alpha$ and $\beta$ qualify as formulas, then C$\alpha$$\beta, K\alpha$$\beta$, E$\alpha$$\beta and B\alpha$$\beta$ qualify as formulas.

Now, we will suppose that C, K, E, and B qualify as truth-functional, meaning that if we input truth values for p and q, C(p,q), K(p,q), E(p,q), and B(p,q) all indicate some sort of truth value. Letting 0 stand for falsity, and 1 for truth, the following "matrix" captures those four functors semantics (as I understand things, this actually comes as instantiation of their matrix, but this will do).

 K|  0  1| B|  0  1| E|  0  1|  C|  0  1
---------------------------------------
0|  0  0| 0|  0  1| 0|  1  0|  0|  1  1
1|  0  1| 1|  0  1| 1|  0  1|  1|  0  1


For defining the ⇒ functor we want a few things:

1. If ⇒$\alpha$$\beta and \alpha qualify as a theorem or tautology, then \beta qualifies as a theorem or tautology (a rule of detachment) for our logic. This means we'll have a rule of detachment, and also indicates why I selected the 4 functors above. 2. Tautologies will exist where whatever member of {B, C, K, E} corresponds to the material conditional "⇒". In other words, if we have a formula \alpha such as BBpqBqr which has only symbols for the material conditional and propositional variables (lower case letters), some tautology will exist. We want some formula which will hold true no matter what the propositional variables stand for (a tautology). Or equivalently we want to make sure that some theorem (not in the axiom set) will exist for some axiomization of a logical system with just the symbol for the material conditional as the only connective appearing in formulas, and that our deductive system will come as sound with respect to our semantics. In other words, we want to ensure that our theorems will qualify as tautologies also (completeness I don't see as necessary). 3. We don't want it the case that if ⇒\alpha$$\beta$ holds, that ⇒$\beta$$\alpha$ holds also, at least not in general. Why? Because, we want to say "if p, then q" can come as true but that "if q, then p" might not come as true.

Now, how do we go about picking from {K, B, E, C} for the material conditional "⇒"? Well, suppose we know next to nothing about logic, and we have to figure things out from scratch. What tautology qualifies as the simplest to figure out for a system with only one binary connective? Well, since we know that ⇒ qualifies as binary by the formation rules above, we can infer it might look something like:

⇒pq.

But, none of the connectives above make ⇒pq into a tautology. But, this will work:

⇒pp

⇒pp is not a tautology if we interpret ⇒ as B or K. But, it does qualify as a tautology if ⇒ gets interpreted as C or E. So at this point, we can reason that ⇒ will come as one of C or E. But by condition 3., E won't work.

Since we've eliminated the other three cases it follows that ⇒ is C.

This is known as ex falso quodlibet, i.e. false implies anything:

If $P$ is not true, then $P\Rightarrow Q$ does not allow you to deduce $Q$. Therefore, both $F\Rightarrow T$ and $F\Rightarrow F$ are valid expressions, i.e. true.

Another way to look it is to consider the statement $P \to Q$ to be interpreted as a relation among the truth values of the statements P and Q, meaning "Q is at least as true as P". This interpretation makes sense for the cases where P is false, since any statement, true or false, is at least as true as a known falsehood. It does not require any logical connection between the content of P and Q; only a knowledge of their truth values. It agrees with the requirements of valid deductive reasoning, which attempts to assure that we do not start with true premises and reach false conclusions. This interpretation can also be extended to multi-valued logic, while the definition of the material conditional $( \sim P \lor Q)$ is not so easily extended.

I too had a considerable amount of trouble with this when I first learned about conditional statements. The accepted answer is an excellent one, but I recently thought of another way of explaining it that I believe is worth sharing here.

Ignoring for a moment the cases where $$P$$ is false, we are left with only the first two rows of the truth table:

$$\begin{array}{ c | c || c | } P & Q & P\Rightarrow Q \\ \hline \text T & \text T & \text T \\ \text T & \text F & \text F \\ \end{array}$$

It can be shown using only these rows that $$P \Rightarrow Q$$ is equivalent to the statement "Whenever $$P$$ is true, $$Q$$ is also true".

At first glance, it seems like this means that it shouldn't matter whether $$Q$$ is true or false when $$P$$ is false. We must look closely at exactly what it means for a predicate to be true or false.

First, note that $$P$$ and $$Q$$ are often not single statements, but functions defined over a set of statements to $$\{\mathbf{T}, \mathbf{F}\}$$ (and technically, even if $$P$$ is a single statement, it can be thought of as a function defined over a set that contains only that statement.) If $$P$$ and $$Q$$ are defined over $$U$$ and $$V$$, respectively, then $$P \Rightarrow Q$$ is shorthand for the following:

$$\forall x \in U \forall y \in V: P(x) \Rightarrow Q(y)$$

Think about exactly what this means. For $$P \Rightarrow Q$$ to be true, $$P(x) \Rightarrow Q(y)$$ has to be true for all $$x \in U$$ and all $$y \in V$$. Only one case of $$P(x) \Rightarrow Q(y)$$ being false is required for $$P \Rightarrow Q$$ to be false. With this in mind, consider what the value of $$P \Rightarrow Q$$ when $$P$$ is false entails.

First, we will select an arbitrary element from $$U$$ and an arbitrary element from $$V$$, and denote them $$r$$ and $$s$$, respectively. Assume $$P \Rightarrow Q$$ is true and $$P(r)$$ is false. $$P(r) \Rightarrow Q(s)$$ being true does not contradict the truth of $$P \Rightarrow Q$$, as it is still true that

$$\forall x \in U \forall y \in V: P(x) \Rightarrow Q(y)$$

However, if $$P(r) \Rightarrow Q(s)$$ is false, we have found one case where $$P(x) \Rightarrow Q(y)$$ is false. As mentioned above, this is all that is required to contradict the assumption that $$P \Rightarrow Q$$ is true.

In conclusion, $$P \Rightarrow Q$$ being true when $$P$$ is false is consistent with the definition of conditional statements, while $$P \Rightarrow Q$$ being false when $$P$$ is false is not.

Short answer: These cases should be true, because the truthness of the conditional proposition in the material implication could be unknown (for comprehension read further; I don't use predicate-logic for explanation).

# Introducing summary

The difficulties of understanding the material implication are caused by two things:

a) The intuitive way of alwalys trying to recognise causal relationships.

b) Insufficient expression of the material implication in common natural language.

I will explain the material implication in the follwing using two steps:

1. Give you an analogy to get a conceptual understanding of the way a material implication connects two propositions.

2. Show you why it is reasonable to have the implication always to evaluate as true if the conditional proposition is false.

## Step 1: Conceptual understanding of the connection expressed by the material implication

Try to think of the material implication in some perspective as an "one way correlation" ("correlation" like e.g. correlation in statistics). Having two corresponding such "one way correlations" you have a "correlation" (biconditional).

E.g.

$I:$ The banana imports are high.

$B:$ Many babies are born.

Let's assume that both propositions are true. The "one way correlation" "$I\implies B$" says that if the banana imports are high, (at the same time) many babies are born. When we look at the collected data, we see that the banana imports are high and that many babies are born, thus the "one way correlation" is true.

Having two corresponding "one way correlations" ($I\implies B$ and $B \implies I$) we have a "correlation" $I\iff B$. So, when we look at the data and see both has happend (many babies are born and the banana imports are high) we can say, that the "correlation" is true. As you see, obviously the banana imports don't have anything causal to do with the birthrate of babies, still the correlation here is true.

Don't think too deeply about this analogy (as it is difficult to think about the "one way correlation"). Just use the ideas you got on the first impression.

Learning-Question: What is the common feature/ the link/the connection, that couples $I$ and $B$?

It is the "same time". The math universe does not have "different times", so everything is always at the "same time". (If you think, there are in some perspective "different times" (read only further if you think that, otherwise the follwing will probably confuse you), then they are just created by you imagining different scenarios. But the logic systems does not have these scenarios, right now in some perspective all truth values are already set [although we maybe didn't found them out yet]. )

Visualisation of the propositional "layer" (I yet don't have enough reputation for embedding). An proposition $P\implies Q$ is only on the propositional "layer"(just a proposition as $\neg P \lor Q$ - it "just" happens to be that $Q$ is true if $P$ is true)[the problem in understanding is that we have to decide the truth values for the implication in general and not on specific cases]. In some sense the causal layer here is the modus ponens.

## Step 2: Evaluating an implication as true if the conditional proposition is false

Now after understanding the nature of coupling expressed by the material implication (and therefore by the propositional abstraction) I add some information for the core of your question. And for that I will use a real world example, because that's what we're used to understand. Because our understanding always tries to see causal relations, I will use an example which does have a (real world) causal relationship. But note that this is only in some perspective a "special case" of an material implication as it is additionally of having a propositional relation also has an real world causal relationship [raining and wetness is a "one way correlation" and has a causal relationship]. But that is what we often do: we look at a visual/concrete special case to understand concepts we project on the general case, e.g. if you try to get insights of a field, you might look at the special case of $\mathbb{R}$ on a graph).

$R:$ It rains on the grass of my new friend's garden.

$W:$ The grass of my new friend's garden is wet.

And $L1$ (in the following also named the law 1): $L1:\iff ( R \implies W )$

Assume $R$ and $W$ are true: Now we say "if it rains, the grass gets wet". This proposition $L1$ is clearly true. True enough, but carefully read the sentence again and notice what you are really doing. Probably you are rather evaluating the wetness of the grass ($W$) than you are evaluating the law 1 itself ($L1$).

Now let's say it, it does not rain ($\neg R$ is true) and the grass is wet ($W$ is true). What would you intuitively think, what that says about the truthness of law 1 ($L1$)? In my view, it says pretty nothing about the truthness about law 1 ($L1$), except that law 1 ($L1$) is pretty useless because when it (always) does not rain, what do I care about a law that says something about when it is raining (still it might be true).

You probably know greenhouses/glasshouses. Obviously when there is a big glasshouse above the whole grass (grass of new friend), the grass won't get wet. The problem with that is, we don't know (yet) if there exists a big glasshouse at my new friend's garden (because I yet wasn't in his garden).

$H:$ There is a glasshouse above the grass (we still don't know if it's there or not, I wasn't yet at my new friend's garden).

Additionally we modify $R$ beeing dependent of $H$ (and correspondig $L1$ ):

$R(H):$ If there is a glasshouse, it does not rain on the grass. If there is no glasshouse, it does rain on the garden. ($R(H):\iff \neg H$)

$L1(H):\iff R(H) \implies W$

Now assume it is (always) raining (not necessarly on the grass). Now we have two scenarios. At first we assume there is an glasshouse ($H$ is true), so it is not raining on the gras ($R(H)$ is false). Our law 1 ($L1$) now is pretty useless as it was before, because it tells us something about what happens, when it's raining on the grass (the truthness of law 1 is "irrelevant"). Now we assume there is no glasshouse above the garden, hence it is raining on the gras ($R(H)$ is true). When we now see(while raining on the grass), that the grass is wet, our law is true.

So, assume the law is true. This might be irrelevant to the first scenario with $R(H)$ beeing false ($false\implies true/false$), but the law itself (although pretty useless in that case) is true (as is the material implicaton corresponding to that law).

Assume that the law is false. This is also irrelevant to the first scenario of not raining on the grass ($false\implies true/false$, we cannot say anything about that because our law has nothing to do with that cases). It is just false, because when we see it raining on the grass, the grass is not wet.

As we are handling the law as a junctor, we must decide what truth value the law has, if the conditional proposition is false and the law doesn't say anything to us. Just imagine we would say the law is false in that two cases ($(false \implies true/false)$ is always false). Then $(R(H) \iff W):\iff R(H) \impliedby W \land R(H) \implies W$ would be false if $R(H)$ and $W$ are false. In other words: It rains on the grass if and only if [exactly when] the grass is wet would be false if it is not raining on the grass and if the grass is not wet. We don't want to have that.

I will probably improve this answer in the future as I'm still not completeley pleased and as a good explanation here is important to me.