Couldn't we have defined the material conditional differently?

Question

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?

Cheers.

Answer 1

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.

Answer 2

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.

Answer 3

... 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 $P \Leftrightarrow Q$.

Answer 4

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.

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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.

Answer 5

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:

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

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EDIT:

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.