truth tables and direct proof

Question

I recently reread about truth tables and I wanted to check whether I understand some things correctly. Essentially, when $p$ and $q$ are formulas of our language, then there is a symbol $p \implies q$ for which one can define the following truth table.

\begin{array}{c|c|c} p & q & p \implies q \\ T & T & T \\ F & T & T \\ T & F & F \\ F & F & T \end{array}

Thus, especially, if $p$ and $q$ are true, then $p \implies q$ is correct as well.

My questions: $(1)$ Is the truth table actually a definition/convention? That is, am I defining the truth values of the symbols there? If so, one could in theory define them arbitrarily right? The problem that would then arise, however, is that our language might/will not be a good model for our reasoning. So the reason why one defines them the way they are is to match what we expect to be the case when reasoning informally, right?

$(2)$ Can I write something like $(p \wedge q) \implies (p \implies q)$ to denote the observation above?

$(3)$ I wondered how one can prove that the method of direct proof actually gives an implication. That is, when assuming $p$ and deducing $q$, why can I deduce $p \implies q$?

$(4)$ Are there deduction rules that use a wrong premise? I know that there is the rule that when $p$ is true and $p \implies q$ is true that one can deduce $q$, which is what one intuitively would think. However, since on defines $p \implies q$ to be true when $p$ is false, are there any deduction rules that are built on this case or are the second and fourth columns just never used? If they are never used, aren't they quite useless and only there to satisfy the claim that they have a truth value?

An attempt to answer $(3)$ might be: If we assume $p$ and can deduce $q$, then we know that whenever $p$ is true, $q$ must also be true. Thus, we can get the cases that both are true, or that $p$ is false. In all of these cases, the implication is true. This appears to be rather informal to me, is there a more formal way to do this?

Answer 1

$(1)$ Is the truth table actually a definition/convention? That is, am I defining the truth values of the symbols there? If so, one could in theory define them arbitrarily right? The problem that would then arise, however, is that our language might/will not be a good model for our reasoning. So the reason why one defines them the way they are is to match what we expect to be the case when reasoning informally, right?

Yes! All correct: Yes, it is a definition, and yes we can define operators any which way we want. However, in trying to capture the natural 'if ... then ...', the truth-table as defined for the $\to$ seems to be the best candidate. It's not perfect (see Paradox of Material Implication) ... but it has some very good features that do seem to match our conceptual understanding of 'if .. then ..' statements

For example, we have that $P \to P$ is always true. We also have that if we have $P \to Q$, and we have $P$, then we have $Q$. We also have that $P \to Q$ is equivalent to $\neg Q \to \neg P$. We have that if we have $P \to Q$ and $Q \to R$, then we have $P \to R$. So these are all things that totally coincide with our intuitions about the conditional, and there are many more examples like this.

$(2)$ Can I write something like $(p \wedge q) \implies (p \implies q)$ to denote the observation above?

I think what you really want to write is $P \land Q \Rightarrow P \to Q$. That is: you want to say that the statement $P \to Q$ is true whenever the statement $P \land Q$ is true, i.e. that $P \land Q$ logically implies $P \to Q$. And we typically use $\Rightarrow$ for the meta-logical symbol of logical implication, and the $\to$ for the logical symbol of material implication. (indeed, note that I was using $\to$ in the previous point ... I really don't like to see $\Rightarrow$ being used for the material implication, even though several texts and websites.

If you don't like to use meta-logical symbols, then what you can do is to say that the logic statement $(P \land Q) \to (P \to Q)$ is a tautology: a statement that is always true.

By the way, note that $P \to Q$ isn't just True when both $P$ and $Q$ are true ... it is also automatically True just whenever $Q$ is True! So, we also have $Q \vDash P \to Q$. Or, if you prefer, that $Q \to (P \to Q)$ is a tautology! And that is actually a pretty good example of where the truth-table for the $\to$ is maybe not such a good match for the English conditional ... we typically don't consider any conditional to be True just because its consequent is True. So this is one of the Paradoxes of Material Implication. These Paradoxes arise because the English conditional isn;t truth-functional .... so no truth-table (which inherently assume truth-functionality) is perfect! But again, the $\to$ is as good as it gets.

$(3)$ I wondered how one can prove that the method of direct proof actually gives an implication. That is, when assuming $p$ and deducing $q$, why can I deduce $p \implies q$?

Typically, when you assume $P$ and deduce $Q$, you do this within the context of other given statements. So, let's say that $\Gamma$ is the set of statements that you have, that you now assume $P$, and then derive $Q$. What this means is that $P$, together with the given set of statements $\Gamma$, implies $Q$. This we typically write as $\Gamma, P \vDash Q$. When you now ask whether this allows us to infer $P \to Q$, you are effectively asking whether $P \to Q$ is a logical consequence of $\Gamma$ alone, i.e. whether $\Gamma \vDash P \to Q$. And yes, that is indeed the case: Given that we have that $\Gamma, P \vDash Q$, we know that it is impossible for $Q$ to be False if all statements in $\Gamma$ and statement $P$ are all true. But that means that it is impossible for $P$ to be True and $Q$ to be False if all statements in $\Gamma$ are True. But that means that $P \to Q$ cannot be False when all statements in $\Gamma$ are True. Hence we have $\Gamma \vDash P \to Q$. This is the formal semantical justification of a direct (or conditional) proof.

Note that if you don't have any given statements, then $\Gamma$ is the empty set. But by the above reasoning, that means that if you assume $P$, and derive $Q$, then $P \to Q$ will always be true, i.e. it will be a logical tautology.

$(4)$ Are there deduction rules that use a wrong premise? I know that there is the rule that when $p$ is true and $p \implies q$ is true that one can deduce $q$, which is what one intuitively would think. However, since on defines $p \implies q$ to be true when $p$ is false, are there any deduction rules that are built on this case or are the second and fourth columns just never used? If they are never used, aren't they quite useless and only there to satisfy the claim that they have a truth value?

I am not so sure I understand the beginning of your question, but I can answer the last part. No, those values in row 2 and 4 are not useless, and it is easy to come up with deductive rules whose validity relies on those very rows being set to True as they are.

First, consider the following rather trivial inference:

$\therefore P \to P$

That is, from nothing we infer $P \to P$. Is that a valid inference? Sure! Because $P \to P$ is true no matter whether $P$ is True or False, this should work. But note: if $P$ is False, then we are dealing with row 4 of the truth-table. So, good thing we defined that as True! (I told you the truth-table has useful features!)

If you don't like this inference, then how about this one:

$P \to Q$

$\therefore, \neg Q \to \neg P$

This we know as Contraposition (actually, the equivalence between these two statements is known as Contraposition, but as an inference from one to the other it should certainly work as well), and again reflects a perfectly intuitively valid inference. OK, but now consider when we set $P$ and $Q$ both to True. Then, by row 1, $P \to Q$ is True, and hence $\neg Q \to \neg P$ should be True as well. But both $\neg P$ and $\neg Q$ are False, meaning that we are relying on row 4 to set $\neg Q \to \neg P$ to True. So this inference too relies on row 4 having the truth-value that it does (or at least, that row 1 and row 4 have the same truth-value, but since you are apparently ok with row 1, then yiou should now be ok with row 4 as well)

Next, consider the following pattern, which is known as Weakening the Consequent:

$P \to (Q \land R)$

$\therefore P \to Q$

Again, this should be a valid inference pattern, right? That is, if we know that both $Q$ and $R$ are true when $P$ is True, then certainly $Q$ by itself should be true when $P$ is True.

OK, but now consider the following case: Set $P$ and $R$ to False, and $Q$ to True. This means that $Q \land R$ is False, and hence $P \to (Q \land R)$ is True, because of row 4 of the truth-table. OK, but then that means that $P \to Q$ should also be true ... and for that we use row 2!

Answer 2

(1) Right. (2) Right.

(3) An informal way to think about it (similar to what you suggested) is: If you assume $p$ and deduce $q$, then you've shown that we can't have $p$ true and $q$ false together. This rules out the only row of the truth table that makes $(p\implies q)$ false, so you can conclude that $(p\implies q)$ is true. You could formalize this using the axiom $p\lor\lnot p$ and a corresponding proof by cases.

(4) Think about how the explanation above relies on the complete truth table for $(p\implies q)$. If there were a row with $p$ false and $(p\implies q)$ false, then assuming $p$ and deducing $q$ would not be sufficient to conclude that $(p\implies q)$ is true, so $\implies$ would fail to model the intuitive notion of conditional implication.

Here's another argument. We want to be able to compose propositional connectives like $\implies$ with quantifiers like $\forall$. Consider the following true sentence:

For all real numbers $x$, if $x>10$ then $x>0$.

For the whole $\forall$ sentence to be true, the inner $\implies$ sentence ("if $x>10$ then $x>0$") must be true no matter which number we substitute for $x$. For example, we need "if $3>10$ then $3>0$" to be true. The standard truth table for $\implies$ arises from these considerations.

Answer 3

(1) Is the truth table actually a definition/convention?

Truth tables are often used to define the standard logical operators, but they can also be formally derived from "first principles", i.e. from basic rules of inference. The truth table for implication, for example, is usually given as:

This table can be effectively derived by proving the following theorems of propositional logic, each theorem corresponding to a line in the table respectively:

1. $A\land B \to (A\to B)$
2. $A \land \neg B \to \neg(A\to B)$
3. $\neg A \land B \to (A\to B)$
4. $\neg A \land \neg B \to (A\to B)$

Using a form of natural deduction, we can formally derive (1) as follows

The rule of inference used for each line is given, e.g. line 1 uses the Premise Rule, and line 3 uses the Split Rule applied to line 1.

(That also answers your next question.)

Similarly, (2) can be derived as follows:

The derivations of (3) and (4) are left as an exercise.