I've been mulling this over lately, and I can't seem to understand why exactly the material conditional wasn't defined in a completely different way.

Obviously, the motivation behind our current definition of the material conditional is, amongst others, so that we don't end up with a definition that makes most restricted quantifiers impossible to work with. For example, take two triangles. A triangle is called equilateral if the lengths of its three sides are the same, whereas a triangle is defined to be isosceles if two of its three sides have the same length. We know then, by definition, that any equilateral triangle is also an isosceles triangle.

With this in mind, we can set up the following implication where $P(T)$ and $Q(T)$ are open sentences over the domain S of all triangles: $$P(T): T \; \textit{is equilateral.} \; \text{and} \; Q(T): T \; \textit{is isosceles.}$$

$$P(T) \implies Q(T)$$

This implication should then be required to hold true for all triangles $T$ that are equilateral, so we could write:

$$\forall \, T, \, P(T) \implies Q(T)$$

but we run into the problem of "what if T isn't equilateral?" If we choose to define the truth values of the material conditional any other way than we currently have, say by making the implication false whenever $P(T)$ is false, we'd end up with a false implication. But surely, our implication (which we know to be true by definition) can't be proven wrong by looking at triangles that the implication doesn't even deal with in the first place. Therefore, we must require this implication to be vacuously true.

However, couldn't all of this have been avoided by simply stating:

$$\forall \, T: \textit{T is equilateral}, \, P(T) \implies Q(T)$$

and assigning a false truth value to the material conditional whenever its antecedent is false? That way, we'd avoid situations like:

$$\forall x \in \mathbb{R}, \; x>2 \implies x^2>4$$ $$x=-1$$ $$\therefore (-1)^2>4$$

...situations which we know to be untrue.

I'm probably not really getting the big picture here, but it seems to me that a simple specification of the properties of a quantified statement fix the issue of quantifier incompatibility. Could somebody enlighten me?


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    $\begingroup$ I don't really get the way you are formalizing: Wouldn't $\forall T:T is equilateral,P(T)\Rightarrow Q(T)$ be saying that "Given any triangle T such that T is equilateral, if T is equilateral then T is isosceles"? Why do you condition on something that is already true in the domain of discourse? Also, from $x>2\Rightarrow x^2>4$ and $x=-1$ because $x=-1$ is not greater than 2, the "whole" implication holds true for that value but not the consequent alone ($\neg(x^2>4)$). $\endgroup$ – Jonathan Julián Huerta Jun 8 '15 at 21:54
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    $\begingroup$ You may enjoy Angelika Kratzer's paper Conditionals which discusses the material conditional and why it is interesting. It comes to the conclusion that the mathematics uses the material conditional because in a certain sense it is the weakest possible form of the conditional. $\endgroup$ – MJD Jun 9 '15 at 16:36

Your inference is wrong.

If we start with the reasonable law :

$∀x \in \mathbb R,x > 2 \to x^2 > 4$

and we instantiate it with $x=-1$, we get :

$-1 > 2 \to (-1)^2 > 4$.

But we cannot infer the false : $(-1)^2 > 4$ because, in order to do it, we have to use modus ponens, and to do this we have to assert the false : $-1 > 2$, and we have no reason to assert it.

  • $\begingroup$ I see, you're right. But then why is the material conditional defined the way it is? I don't get it! $\endgroup$ – Ius Klesar Jun 9 '15 at 14:52
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    $\begingroup$ @Luke - exactly because we want modus ponens to work ... If the conditional $p \to q$ is true and the antecedent $p$ is true, then we want to conclude with the consequent. This fact account for the case when $p$ is true. If we set $p \to q$ false when both are false, we have to "break" the equivalence between $p \to q$ and $¬q \to ¬p$. In conclusion, the only "debatable" case is FALSE $\to$ TRUE... but if we change only this one to false what we get is the bi-conditional : ↔. $\endgroup$ – Mauro ALLEGRANZA Jun 9 '15 at 15:14
  • $\begingroup$ Yes, but isn't the bi-conditional defined based on the material conditional? So aren't you essentially referring back to our current definition of the conditional to demonstrate the necessity to define the conditional that way? Besides, why is the equivalence between $p \implies q$ and $\lnot q \implies \lnot p$ so important? $\endgroup$ – Ius Klesar Jun 10 '15 at 16:38
  • $\begingroup$ @Luke - no, forget about bi-conditional and call it equivalence. It is quite "natural" to say that two sentences are equivalent when they have the same truth-value; thus, no material conditional is involved here. $\endgroup$ – Mauro ALLEGRANZA Jun 13 '15 at 10:34
  • $\begingroup$ @Luke : if $p \to q$ is false when both are false, we have that its contrapositive ; $\lnot q \to \lnot p$ will be true (because now both antecedent and consequent are true). But again, our "intuition" says that if we assert that : "if I'm more than twenty years old, then I was born earlier than 1995", for sure I can assert : "if I was born not earlier than 1995, then I'm not more than twenty years old". $\endgroup$ – Mauro ALLEGRANZA Jun 13 '15 at 10:38

... assigning a false truth value to the material conditional whenever its antecedent is false?

Let's approach $P \Rightarrow Q$ as a truth functional connective. I assume we agree that if $P$ and $Q$ are true, then $P \Rightarrow Q$ is true. Also, we probably agree that if $P$ is true and $Q$ is false, then $P \Rightarrow Q$ is false.

Using that agreement, and the quoted statement, we would find that the truth table for $P \Rightarrow Q$ is the same as the truth table for "$P$ and $Q$". But nobody thinks that "if $P$ then $Q$" should have the same meaning as "$P$ and $Q$"! So the quoted statement above can't work if we want to capture "if $P$ then $Q$".

It turns out that, based on basic reasoning about implication, we should have the following three lines of the truth table:

P  Q  P => Q
T  T   T
T  F   F
F  F   T

That only leaves one missing line: when $P$ is false and $Q$ is true. If we let $P \Rightarrow Q$ be false in that case, we obtain the following table:

P  Q  P => Q
T  T   T
T  F   F
F  F   T
F  T   F

But that table would say that $P \Rightarrow Q$ is the same as $P \Leftrightarrow Q$. Again, nobody thinks that "if $P$ then $Q$" means the same as "$P$ if and only if $Q$".

The only other option is the actual table for material implication. So, rather than being arbitrary, the table for material implication is the only choice that matches the three unambiguous lines of the truth table and does not make $P \Rightarrow Q$ the same as $P \Leftrightarrow Q$.

  • $\begingroup$ But isn't that some kind of circular reasoning? The biconditional is defined based on our current definition of the implication, so arguing that the implication can't be any different from how it is now or else it would clash with the biconditional strikes me as circular reasoning. On a different note; how exactly did you arrive at the first truth table? $\endgroup$ – Ius Klesar Jun 10 '15 at 16:05
  • $\begingroup$ The first truth table contains the three lines that are relatively uncontroversial about a truth-functional connective for "if P then Q". If P and Q are both true, or both false, then the implication should hold; if P is true and Q is false the implication should not hold. The fourth row is the one that causes the most consternation. As for circularity, even before we study formal logic we know that "if N is a multiple of 4 then N is even" is not the same as "N is a multiple of 4 if and only if N is even". $\endgroup$ – Carl Mummert Jun 10 '15 at 16:45
  • $\begingroup$ Yes, we know they're not the same, but before we study formal logic we won't know what their truth values are either. On top of that, I still don't understand why that truth table is the way it is. If P and Q are both false and the implication still holds, I don't really consider that uncontroversial. As for the 4th row; you still haven't explained why it was defined that way? $\endgroup$ – Ius Klesar Jun 10 '15 at 16:51

The one rule that we all agree must work is modus ponens: if we have $p \Rightarrow q$ and $p$ then we must have $q$. This means that $p \Rightarrow q$ must imply $\neg (p \wedge \neg q)$. The question for us is whether this is strong enough.

To help us decide whether that is strong enough, we look at trying to extend our definition of the conditional to first order logic. That is, we turn to statements of the form

$$(\forall x \in U) \: P(x) \Rightarrow Q(x) \quad (1)$$

for predicates $P,Q$ and a domain of discourse $U$. (I ask that you do not bring up set theory issues here.) I think it is reasonable to assert that this should be equivalent to

$$(\forall x \in \{ y \in U : P(y) \}) \: P(x) \Rightarrow Q(x). \quad (2)$$

That is, the members of $U$ which don't satisfy $P$ should not have anything to do with the truth of our statement. This is one point that could possibly be disputed.

The other main philosophical assumption is that in classical logic, every well-formed statement is either true or false. This is the other main point which can be disputed. (This point really is treated differently in some nonclassical logics, such as intuitionistic logic.)

Given these two assumptions, $P(x) \Rightarrow Q(x)$ must be true whenever $P(x)$ is false, because otherwise there would exist situations where (1) is false and (2) is true. Restricting attention back to propositional logic, we get the classical definition of the material conditional.

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    $\begingroup$ @Luke You are asking about the first of my two assumptions, i.e. that (1) and (2) are equivalent. In my answer I assert that this is natural. Maybe an example will help with that. Let's say $U$ is the real numbers, $P(x)$ is defined as "$x$ is an integer", and $Q(x)$ is defined as "$x^2$ is an integer". Do you agree that we should have $(\forall x \in U) \: P(x) \Rightarrow Q(x)$? In natural language this should mean "if $x$ is an integer then $x^2$ is an integer". $\endgroup$ – Ian Jun 10 '15 at 15:48
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    $\begingroup$ @Luke (Cont.) If you accept this, then given the second assumption (that $P(x) \Rightarrow Q(x)$ has to have some truth value), $P(1/2) \Rightarrow Q(1/2)$ must be true. Similarly $P(\sqrt{2}) \Rightarrow Q(\sqrt{2})$ must be true. Yet in the first case $P$ and $Q$ both fail, while in the second case $P$ fails and $Q$ holds. $\endgroup$ – Ian Jun 10 '15 at 15:48
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    $\begingroup$ @Luke What do you think about my example in the previous two comments? $\endgroup$ – Ian Jun 10 '15 at 16:21
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    $\begingroup$ @Luke Because of the second assumption: the statements "$x$ is an integer" and "$x^2$ is an integer" are well-formed for any real number $x$, so all of them must have some truth value. So under the second assumption you cannot simply dodge the issue, you have to decide whether $P(1/2) \Rightarrow Q(1/2)$ is true or false. If we chose false, then the formalization of "for any real number $x$, if $x$ is an integer then $x^2$ is an integer" would be false. That seems awfully problematic to me. $\endgroup$ – Ian Jun 10 '15 at 16:38
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    $\begingroup$ @Luke One somewhat abstract place where this is important is when we prove a theorem in the form of an implication, where we do not know whether the hypothesis is true or false. For instance, there are a lot of theorems in modern number theory that say "if the Riemann Hypothesis is true, then ..." $\endgroup$ – Ian Jun 10 '15 at 16:40

I think it is illuminating to approach it from the opposite direction.

Suppose we already know we want universal quantifiers (those are easy to motivate, I think), and that we also want to express the idea "everything that satisfies $P$ also satisfies $Q$"). It turns out that exactly this meaning can be expressed as $$ \forall x\, [ P(x) \triangleright Q(x) ] $$ if $\triangleright$ is a truth-functional connective with the truth table

$$ \begin{array}{cc|c} P & Q & P \triangleright Q \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \end{array} $$ Since we want to have quantifiers anyway and truth-functional connectives are easy to understand and manipulate, this is a wonderful plan! We just need to decide on a name for our connective.

For better or worse, mathematics has decided to use the symbol $\rightarrow$ (or $\Rightarrow$ or $\supset$) to write this connective and to pronounce it with the words "if ... then ...".

But it's really not a matter of "what should the right meaning of the words 'if' and 'then' be?", but instead "what should we call the thing with such-and-such truth table?".

If somehow, we all are convinced that giving "if ... then" this meaning is so confusing and illogical that it needs to be changed, we would still keep using a connective with the above truth table. We would just call it something different.

It would be technically possible to fuse the two constructions together and say that $\forall x$ must always appear with two formulas: a condition and a conclusion, so $\mathop{\forall x}\limits_{P(x)} [Q_x]$ means what we're currently writing $\forall x[P(x)\to Q(x)]$. Then if we ever needed a quantifier without a condition we could write it as $\mathop{\forall x}\limits_{x=x} Q(x)$.

But this would make things strange whenever we want to quantify over two variables -- for example $\forall x \forall y (x<y\Rightarrow P(x,y))$ would now need to be written $\mathop{\forall x}\limits_{x=x} \mathop{\forall y}\limits_{x<y} P(x,y)$, where the dummy condition on $\forall x$ doesn't exactly make things easier to read.

And generally we want to split complex constructions into simpler ones whenever we can -- that makes all of the theory simpler. So simply because we can split the quantifier-with-condition into a universal quantifier and a truth functional is enough reason to do it.

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    $\begingroup$ Also I think your rejection of OP's requirement of bounded quantifiers is too pessimistic. There are logics and mathematical contexts that require all quantifiers to be bounded, so that "$\forall x"$" is forbidden and only "$\forall x\in S$" is allowed. $\endgroup$ – MJD Jun 10 '15 at 14:39
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    $\begingroup$ Indeed I think some people would argue that unbounded quantifiers are logically incoherent, and when we omit the bounds it's only by way of a convenient abbreviation when the bound is implied by the context. $\endgroup$ – MJD Jun 10 '15 at 14:42
  • $\begingroup$ @MJD: Sure, there are contexts that care about bounded quantification -- but in those contexts you generally want tighter control on the shape of the bounds than just "insert some property here", so having a three-place quantifier as the general construct is not going to make it appreciably easier to define bounded quantification. At least not if the general construct is flexible enough to subsume our current use of logical implication. $\endgroup$ – Henning Makholm Jun 10 '15 at 14:43
  • $\begingroup$ I think that is a result of OP's unfamiliarity with how things are actually done. Had they more experience, they might have suggested, instead of $$\underset{\text{$T$ is triangular}}{\forall T.} P(T)\implies Q(T)$$ something like $$\forall T\in{\mathscr T}. P(T)\implies Q(T),$$ where $\mathscr T$ is the set of triangles. We don't do the first, but we do the second, and I imagine OP is unaware of this, and might view it as being sufficiently similar to his or her idea that the question is resolved not as "we don't" but as "we do"—except for the part about redefining the material conditional. $\endgroup$ – MJD Jun 10 '15 at 15:08
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    $\begingroup$ @Luke: There are only 16 possible truth tables for a truth function of 2 operands; it is not difficult to check that each of the 15 others don't give the desired functionality when put under a quantifier (in the sense that you either can have a situation where all $P$s are indeed $Q$s but $\forall x[P(x)\rhd Q(x)]$ is nonetheless false, or you can have a situation where some $P$s are not $Q$s, yet $\forall x[P(x)\rhd Q(x)]$ evaluates to true). $\endgroup$ – Henning Makholm Jun 10 '15 at 17:06

Consider the statement: If it is raining, then it is cloudy.

Symbolically: $Raining\implies Cloudy$

What does this mean? First, what it does not mean: It does not mean that rain somehow causes cloudiness, or that cloudiness somehow causes rain. Neither is the case. It means only that it is not both raining and not cloudy.

Symbolically: $\neg[Raining \land \neg Cloudy]$

There is no suggestion of causality -- of one thing causing another -- or of time-based data about a historical relationships between raining and cloudiness.

Everything about material implication -- the truth table for $\implies$, and the detachment and contra-positives rules -- all make sense when you understand this. Simply stated, the rule is:

$A\implies B\equiv \neg[A\land \neg B]$


The Truth Table

$\begin{array}{c|c|c|c} A&B&A\implies B\\ \hline T&T&T\\ T&F&F\\ F&T&T\\ F&F&T \end{array}$

The Intuition

In the first row, we have $A\implies B$ and $A$ being true. Of course, $B$ must also then be true, as indicated.

In the second row, we $A$ being true and $B$ being false. Then $A\implies B$ must be false, as indicated.

Note that when $A\implies B$ is true and $A$ is false, as you would expect, $B$ is indeterminate, i.e. $B$ could be either true or false, as indicated in the the last two rows.

We also need $A\implies B$ to be true when $A$ is false to prove vacuous truth, an essential method of proof in mathematics.

  • $\begingroup$ So the only reason that the conditional is the way it is, is because it's logically equivalent to not(A and not B)? That's it? No profound reason as to why it was defined that way? $\endgroup$ – Ius Klesar Jun 9 '15 at 7:40
  • $\begingroup$ That is correct. I think you will find that this definition works really quite well in writing mathematical proofs. It may seem counter-intuitive at first because the if-then construct in everyday usage usually suggests a causal or temporal relationship. In mathematics, there is no cause and effect and no passage of time. It may help to think of mathematics as a static body of knowledge, aspects of which humans have discovered over the millennia. Cause and effect is the realm of science. But even in science, we don't say, if you smoke then you have cancer. We say smoking may cause cancer. $\endgroup$ – Dan Christensen Jun 9 '15 at 15:11
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    $\begingroup$ I think this misses the point, and I think the point here is from Mauro ALLEGRANZA's comment above: We define it this way because that's what makes modus ponens work. $\endgroup$ – MJD Jun 9 '15 at 16:34
  • $\begingroup$ @MJD In mathematical proofs, we also need $A\implies B$ to be true if $A$ is false. It's not just some undesirable side-effect as some would have it; it is essential. $\endgroup$ – Dan Christensen Jun 9 '15 at 16:40
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    $\begingroup$ Okay. I still feel that your answer misses most of the crucial issues, and it doesn't address OP's suggestion either. $\endgroup$ – MJD Jun 9 '15 at 17:13

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