Why are truth tables not circular?

Question

Why are logical truth tables not circular? For example, let’s consider the task of defining the concept of implication. In fact, in using a truth table, we say that each combination of truth values implies another. So we are saying, for example, that if $P$ and $Q$, then $P\Rightarrow Q$, and so on for each possible truth combination. Is really the best way to define “$\Rightarrow$” as $$P\wedge Q\Rightarrow P\Rightarrow Q$$

$$P\wedge \lnot Q\Rightarrow \lnot(P\Rightarrow Q)$$

$$\lnot P\wedge Q\Rightarrow P\Rightarrow Q$$

$$\lnot P\wedge \lnot Q\Rightarrow P\Rightarrow Q$$

Therefore, I feel wary of using truth values to define logic because it seems to explicitly use that same logic in its construction. I suppose that this question further asks: how can logic really be defined, without using logic; for example, if we tried to define logic under a set of axioms, how would we ever be able to derive results regarding the nature of logic; that is to say, logic is the way we move from axioms to implications, how can we make implications about logic itself?

Answer 1

• Your statements are not correct, for you are confusing material implication and logical implication.

P materially implies Q ( symbol $\rightarrow$) iff it is ( factually) not the case that P istrue and Q is false ( or, equivalently, we are factulally not in a case where P is true and Q is false).

P logically implies Q ( symbol $\Rightarrow$) iff it is ( logically) impossible that P is true while Q is false. ( or , equivalently, the case where P is true and Q is false is impossible)

Hence there is no circularity. Truth tables and material implication belong to object language. The conclusions we draw in terms of what logically implies what belong to metalanguage, or to the metalogical level.

• Take for example your first formula : $(P\wedge Q)\Rightarrow (P\Rightarrow Q)$.

It says that $(P\land Q)$ logically implies that " P logically implies Q" ( or, if you prefer, it says that $(P\land Q)$ logically implies that $(P\rightarrow Q)$ is a tautology ).

The following truth table shows that : the fact that $(P\rightarrow Q)$ is not false when $(P\land Q)$ is true does not make of $(P\rightarrow Q)$ a tautology, that is a formula that is true in all 4 possible cases.

What is true, in fact, is that : $(P\land Q)$ logically implies that P materially implies Q,

in symbols : $(P\wedge Q) \Rightarrow (P\rightarrow Q)$. ( The tautology is the central conditional, not the conditional embedded in the consequent).

• Let me show why your sentence is false by a counter example.

$(P\wedge Q)\Rightarrow (P\Rightarrow Q)$

$\equiv$ there is no possible case where ($(P\land Q)$ is true and $(P\Rightarrow Q)$ is false )

$\equiv$ there is no possible case where ( $(P\land Q)$ is true and there is a possible case in which $P$ is true and $Q$ is false)

But consider the case where both $P$ and $Q$ are true. In that case $(P\land Q)$ is true. This possible case certainly exists. But the fact that this case exists certainly does not prevent the possibility of the case where $P$ is true and $Q$ is false. !

• In terms of modal logic, with $\square X$ meaning " it is necessarily the case that X" :

(1) $\square [(P\land Q) \rightarrow ( P\rightarrow Q)]$

( In words : " necessarily ( P&Q implies that, if P, then Q)" )

is correct , but is not the same thing as

(2) $\square [(P\land Q) \rightarrow \square ( P\rightarrow Q)]$

(In words : "necessarily ( P&Q implies that P necessarily implies Q)" ).

Note : on the distinction between material / logical implication, Lipschutz, Outline Of Set Theory (Chapter on the Algebra Of Proposiitons) ( At archive.org).

Answer 2

All of math proceeds by postulating some set of axioms, and then writing out all of the logical conclusions that follow from those axioms. The truth tables just give the rules for what the valid deductive moves are in the game that is being played. You could change the rules, but then you would either be able to derive $P$ and $\neg P$ or there would be true statements you could not deduce in your system that would be true in the standard one, and your system would not be very helpful.