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A lot of people regard the conditional truth tables as being a little dubious. If not dubious, then we can at least say they are different from the non-conditional operators in that they are the only ones that you might describe as 'innocent until proven guilty.' That is, if $P$ is F, $Q$ can be F or T, and the conditional is true.

E.g. If the car's in the driveway ($P$), then the driver is home ($Q$)

If the driver is home and the car's not in the driveway, we can't say that it's not true that the car being in the driveway is conditional on the driver being home, and the statement evaluates as true.

The only time we can say that the statement is false is when the car is in the driveway, and the driver is not home, and we have proof, as it were, that it's not true that the car being in the driveway is conditional on the driver being home, and the statement evaluates as false.

The non-conditional operators don't seem to have that weirdness.

Also, they don't seem to map to most people's idea of 'if-then' statements, and rules about assertability don't fix the issue entirely.

But, of the 16 possible operators, it seems like they all correspond to some sort of natural language coordinator - even if English doesn't have one word for each of them (e.g. exclusive disjunction). So, it seems unlikely that the conditional truth table maps to something that's universally linguistically empty.

Saying all that to ask, are there any non-conditional translations for the conditional operators that you find plausible?

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You are in good company ! because a lot of distinguished logicians moved objections to the "naturalness" of the truth-functional analysis of the conditional connective (see for example : S.C.Kleene, Mathematical Logic [1962 - Dover reprint], pag.9).

For me, the best way to avoid the problem (and solving it) is to remember that you can develop classical sentential logic with (for example) only $\neg$ ("not") and $\lor$ ("or").

In this way, the conditional $P \rightarrow Q$ is only an (useful) abbreviation for $\lnot P \lor Q$.

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The conditional $P\rightarrow Q$ is true:

$$\text{...whenever}\;\underbrace{P \;\text{is false}}_{\large \lnot P}\;\;\underbrace{ \text{ OR }}_{\large\lor} \;\underbrace{Q \;\text{is true.}}_{\large Q}$$

This translates very easily to the equivalent proposition: $$P\rightarrow Q \equiv \lnot P \lor Q$$

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  • $\begingroup$ needs a TU to compensate for the jealous DV! +1 $\endgroup$ – Amzoti Dec 13 '13 at 2:15

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