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Disclaimer: I'm reading Wikipedia.

I think the truth value of F $\Rightarrow$ T should be undefined. Here's why.

Wikipedia says the Principle of Explosion says anything follows from a contradiction, and gives the following proof:

  1. $P \land \lnot P$
    by assumption

  2. $P$
    from 1. by conjunction elimination

  3. $\lnot P$
    from 1. by conjunction elimination

  4. $P \lor Q$
    from 2. by disjunction introduction

  5. $Q$
    from 3. and 4. by disjunctive syllogism

  6. $(P \land \lnot P) \Rightarrow Q$
    from 5. by conditional proof (discharging 1.)

The disjunctive syllogism on step 5 works only if $P$ and $\lnot P$ are mutually exclusive. But of course they aren't.

I can prove $\mathrm F \Rightarrow \mathrm F$ via contrapositivity and $\lnot(\mathrm T \Rightarrow\mathrm F)$ by other means.

So that leaves $\mathrm F \Rightarrow \mathrm T$. It seems to me that there's no imperative way to define its truth value.

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marked as duplicate by Andrés E. Caicedo, Derek Elkins, Henning Makholm logic Mar 26 '17 at 19:45

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    $\begingroup$ If this is a philosophical issue, I don't think that it belongs here. Mathematically, the question of whether $\bot\implies\top$ is "true" or "has truth value T" is settled in classical logic and in intuitionistic logic. In classical logic, the answer is yes, and I think it's the same for intuitionistic logic. Of course there are other logics in which the answer may differ, but still. So you should make the context of your assertion precise. Your argument as to why the proof doesn't work is not mathematical $\endgroup$ – Max Mar 26 '17 at 18:48
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    $\begingroup$ See Paraconsistent Logic. It may be relevant to your remark. And it also seems to be related to philosophy to me. $\endgroup$ – Boris E. Mar 26 '17 at 18:49
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    $\begingroup$ "The disjunctive syllogism on step 5 works only if $P\vee\neg P$ is a disjunction. But it's not!" What? I think you mean something different by "disjunction" than mathematics does. Besides, even if you view inclusive "or" as unnatural, I don't see how you can possibly not conclude $Q$ from "$P$ or $Q$" and "Not $P$." Can you give me an example of two statements $P$ and $Q$ such that at least one of them is true ($P\vee Q$), and $P$ is false ($\neg P$), but $Q$ isn't true? $\endgroup$ – Noah Schweber Mar 26 '17 at 18:49
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    $\begingroup$ @MackTuesday That's not a problem with disjunctions, that's a problem with vague statements: in classical logic, we're only interested in statements that aren't vague in this way (other logical systems do try to handle such statements, with varying degrees of success). Can you come up with an example that doesn't rely on vagueness in this way? $\endgroup$ – Noah Schweber Mar 26 '17 at 19:12
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    $\begingroup$ If you are interested in pursuing logical systems which do handle such statements, and which treat disjunctions and implications with more subtlety, you may be interested in any of the following: intuitionistic logic, relevance logic, paraconsistent logic, modal logic, fuzzy logic. (Listed in no particular order.) But classical logic is designed for a much more restrictive situation (and that's not necessarily a bad thing - I'm a huge proponent of classical logic over the others, for instance). In the context of classical logic, each statement has exactly one truth value - "true" or "false." $\endgroup$ – Noah Schweber Mar 26 '17 at 19:16
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Your edit hits the nail on the head:

The disjunctive syllogism on step 5 works only if $P$ and $\neg P$ are mutually exclusive. But of course they aren't.

What do you mean "of course they aren't"? In classical logic, of course they are.

Statements in natural language have varying degrees of vagueness, which accounts for odd behavior - e.g. as you say in a comment, "The sky is cloudy," being a qualitative statement, could be construed as both true and not true. However, such statements are outside the purview of classical logic, which is designed to handle only precise statements.

(CAVEAT: My terminology of "vague" versus "precise" quite clearly reflects my own classical bias, and those who prefer a non-classical logic might well take issue with it. I do stand by it, and am prepared to defend it, but I want to point out my own bias here for fairness.)

Now, this isn't to say that such statements can't be treated by mathematical logic - which is much broader than classical logic - at all. Each of the following addresses issues around disjunctions, negations, and implications in an attempt to better reflect the nature of natural language:

But that's not necessarily a flaw in classical logic, it just reflects a narrower focus. We wouldn't say that group theory is inferior to magma theory, after all, even though magmas are far more general! And in fact as a classical-ist, I'd argue that classical logic really underlies each of the others and is the "true" logic (but wow is this ever a controversial statement among logicians, so definitely take this as a statement of my own opinion rather than clear unbiased truth).

So I think the takeaway is this: you are looking for a logic which faithfully handles some of the oddities of natural language, which classical logic certainly doesn't. Whether this is a fundamental flaw in classical logic, or a revealed problem with natural language, will ultimately depend on the interpreter (says I: down with natural language! :P).

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    $\begingroup$ One philosophical argument around this issue is the following. Considering that truth values don't really match up with natural language behaviors (I'd argue e.g. that the Liar paradox shows that we can never have a mathematically precise notion of truth values which cover everything we actually use in natural language), we can consider classical logic to be the logic of ideal language for a certain sense of "ideal." A question then is whether mathematical statements are ideal in this way, and if not, which ones are (and ultimately what constitutes a "mathematical statement"). $\endgroup$ – Noah Schweber Mar 26 '17 at 19:35
  • $\begingroup$ We can say $P$ and $\lnot P$ aren't mutually exclusive because we assume $P \land \lnot P$ in step 1. $\endgroup$ – MackTuesday Mar 27 '17 at 3:45

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