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I've been thinking about this qustion quite a long.
It all started reasoning about this sentence "Either I'm very good at foorball or I can't play it".
Now, I know that a truth function is a function that "accepts truth values as input and produces a truth value as output". In propositional logic we use the truth tables of the logic connectives to evaluate the truth-value of complex (or compound ) statements.
Unfortunately(for me), appealing to my intuition Exclusive Disjunction presents a problem. I'll try to analyze the different cases for the above stentence:

  1. (1,0)=1 : I can imagine a world where I'm indeed very good at playing football, so it makes perfect sense to say that the whole sentence is true.
  2. (0,1)=1 : I can Imagine a world where I really can't play football, so, again, it makes perfect sense to say that the whole sentence is true.
  3. (0,0)=0 : I can imagine a world where I'm average at playing footaball, so in this scenario both the statements "I'm very good" and "I can't paly" are false, and so it is evident that the same holds for the whole sentence.
  4. (1,1)=0 : Now here it is my problem. I can't imagine a world where I'm both good at playing football and unable to play it ( It feels constradictory).Since I can't assign the value truth to both statements at the same time, it doesn't make sense to me to pass them as an argument. Since they can't coexist, they can't be co-assigned as arguments to the function! As soon I choose one to be tru the other turns false so I can say nothing in this scenario.

I think, it turned out a bit philosophical. Hope you got my probelm!

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    $\begingroup$ Your problem is not actually with the exclusive or as such, but rather the statements you put into it - namely their interdependence. You would run into the same situation if you used inclusive or. $\endgroup$ – skyking Oct 26 '17 at 12:31
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Notice that what you really end up with here is that two mutually exclusive statements cannot possibly be true at the same time... which makes perfect sense! It does not mean that the $XOR$ is not a truth-function ... it is, and it is exactly because it is that we can rule out such scenarios. If there is any problem here, it is not with the $XOR$ being truth-functional.

OK, but you say that we cannot possible feed an $XOR$ two statements that are mutually exclusive. Sure enough, but that does not stop one from evaluating the $XOR$ of two statements $P$ and $Q$ that are not mutually exclusive. That is, if I say that "John is good at football or at math but not both", then we can evaluate that statement just fine when John turns out to be both good at football and math: in that case my statement would be false, exactly because both parts are true, and thus exactly as dictated by the truth-table definition of the $XOR$.

Finally, let me point out that if you say that 'being really good at football' and 'cannot play football' are mutually exclusive, then that would be a perfect candidate to symbolize using an $XOR$. That is, if we were to use $GF$ for 'I am really good at football' and $CF$ for 'I cannot play football', we can symbolize $GF \ XOR \ CF$ as a fact. And we can use that fact to draw useful inferences. For example, if we know that $CF$ is true, then we can combine that with $GF \ XOR \ CF$ to obtain $\neg GF$. So again, rather than there being a problem, the $XOR$, and how it is truth-functionally defined, does its very job!

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  • $\begingroup$ Hmm, on the contrary $P \mathbin{\it XOR} \neg P$ cannot possibly be false. $\endgroup$ – hmakholm left over Monica Oct 26 '17 at 12:59
  • $\begingroup$ @HenningMakholm Right! :P $\endgroup$ – Bram28 Oct 26 '17 at 14:29
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What mathematicians (and computer engineets/programmers) mean when we say "exclusive disjunction" is certainly a truth function: It is defined by a certain truth table, and that truth table is the thing we really want to get when we use it.

The English words "exclusive disjunction" is just a label we use to talk about the truth table. They're supposed to be somewhat evocative, helping us to remember which operation it is we call by that name, but the natural-language meaning of each of the words does not participate in defining the technical thing they refer to.

Your question appears to be more about whether using XOR is a reasonable way to model the meaning of a particular English sentence. That is a question about how English works, not about mathematics. If the conclusion is (as it might well be) that the meaning English speakers get from hearing the sentence does not match our exclusive-disjuction truth table, the conclusion is not that there is anything wrong with the logical connective, merely that your English sentence happens not to correspond to that particular connective.

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  • $\begingroup$ I need to make a blog post I can point to arguing that teaching logic by reference to English sentences, especially "every day" sentences, is not just ineffective but actually harmful (except maybe if your goal is to be a linguist). One particular aspect of this is the seemingly large emphasis placed on "translating" English statements into logical formulae for no reason that is apparent to me. The only people who ever do anything remotely like this are linguists. $\endgroup$ – Derek Elkins left SE Oct 26 '17 at 12:51
  • $\begingroup$ @DerekElkins I have seen natural language arguments translated into formal language before. For instance, some of Spinoza's arguments have gotten translated into formal language. $\endgroup$ – Doug Spoonwood Oct 26 '17 at 14:29
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You have imagined three worlds, one in which you are really good at football, one in which you can't play football at all, and one in which you're just average at football. I'll call these worlds $G$, $B$, and $A$ (for "good", "bad", and "average"). You've also indicated that these are the only kinds of worlds you can imagine, because you can't be really good at football and simultaneously unable to play football. The statement that either you're very good at football or (exclusive or) you can't play football is true in worlds $G$ and $B$, but false in world $A$. (Note that the same would be true for inclusive or; the two sorts of disjunction differ only when both disjuncts are true, which, in your example, is the unimaginable situation.)

None of this says that exclusive or isn't a truth function. It's defined by a perfectly good truth table. The truth value of your example sentence ("either you're very good at football or (exclusive or) you can't play football") is completely determined by the truth values of the two constituents ("you're very good at football" and "you can't play football").

The fact that some combination of truth values (in this case "true" for both disjuncts) is unimaginable in a particular example makes no difference to what is and what isn't a truth function. As long as the truth table specifies an output for all combinations of inputs (imaginable or not in specific examples), you have a truth function.

(Irrelevant to the logic question but relevant to the vagueness of ordinary English: A football superstar who is temporarily injured might be described as being very good at football yet also unable to play it.)

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You have three predicates indicating a measure of goodness for playing football: unable, average, and very good. Let us call them $U,W,V$.

Then you make a claim that you are unable xor very good.   $U\oplus V$.

This claim is a truth function.   It has two arguments $U, V$ and maps to either true or false.   In terms of $\{\wedge, \vee,\neg\}$ the claim is $(U\vee V)\wedge\neg(U\wedge V)$ among other equivalent forms.

You then further claim that you can only be exactly one from unable, average, or very good; no more and no less.   That they are mutually disjoint, and exhaustive; they partition the possible universes. $$(U\oplus W\oplus V)\wedge\neg(U\wedge W\wedge V) \\ \text{equivalently: } (U\wedge\neg W\wedge \neg V)\vee(\neg U\wedge W\wedge \neg V)\vee(\neg U\wedge\neg W\wedge V)$$

This claim does not prohibit $U\oplus V$ from being a truth function of $U,V$.   It merely a second claim.   If it is true, then it implies that $(U\oplus V)\oplus\neg (U\vee V)$.

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