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I find that I frequently make use of the following claim in different arguments (when I am trying to exhaust all possibilities):

$\forall x \left[ x = c \lor x \neq c \right ]$ for some object $c$.

Why is it that I am permitted to make this claim? In particular, I guess I am asking what definition/axiom of equality allows me to deduce this universal statement. For real numbers, I am familiar with trichotomy, which states that $\forall x \in \mathbb R \left [ x=c \lor x \lt c \lor x \gt c \right ]$ for some object $c$...but the aforementioned universal claim is certainly even more fundamental.

Shaun
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S.C.
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3 Answers3

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Keep in mind that $x \neq y$ is just shorthand for $\neg(x = y)$. So what you're really proving is $\forall x (x = c \lor \neg (x = c))$.

Given an arbitrary $x$, the fact that $x = c$ or $\neg (x = c)$ follows from the law of excluded middle.

Note that in settings where the law of excluded middle is absent, one cannot always assert $\forall x (x = c \lor x \neq c)$. However, there are many cases in which one can assert this.

For example, one can prove using induction and the Peano axioms that $\forall y \forall x (x = y \lor x \neq y)$, where the domain of discourse is the natural numbers.

However, when one is dealing with sets, one cannot assert that $\forall x (x = \{\emptyset\} \lor x \neq \{\emptyset\})$.

This is because if one could assert this, then consider an arbitrary proposition $P$. Let $w = \{x \in \{\emptyset\} \mid P\}$. Then $w = \{\emptyset\}$ if and only if $P$. So $P \lor \neg P$ is equivalent to $w = \{\emptyset\} \lor w \neq \{\emptyset\}$. So asserting $\forall x (x = \{\emptyset\} \lor x \neq \{\emptyset\})$ is equivalent to asserting the law of excluded middle.

Mark Saving
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Suppose otherwise. Then there exists an $a$ in the domain such that $\lnot (a=c\lor a\neq c)$, which means, by de Morgan's laws, $\lnot (a=c)\land\lnot(a\neq c)$. But this means $a\neq c$ and $a=c$, a contradiction.

Shaun
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    this is making use of the principle of non-contradiction, right? Where $\varphi(a)$ is defined as $a=c$...and therefore your final claim reads as $\varphi(a) \land \neg \varphi(a)$. – S.C. Aug 12 '21 at 23:40
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    Yes, @S.Cramer. – Shaun Aug 12 '21 at 23:42
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From your comment for your accepted answer, it's worthwhile to point out that you're still conflating law of excluded middle (LEM) with another law of noncontradiction (LNC).

For LEM:

It is one of the so called three laws of thought, along with the law of noncontradiction, and the law of identity... (p $\lor$ ¬p)

For LNC:

Formally this is expressed as the tautology ¬(p ∧ ¬p)... The law is not to be confused with the law of excluded middle which states that at least one, "p is the case" or "p is not the case" holds.

So under classic propositional logic you learned in school, obviously they're equivalent per DeMorgan, however, in intuitionistic logic LNC still holds while LEM together with double negation have been removed. On the other hand, in paraconsistent logic such as relevance logic inconsistency is tolerated and thus rejects the LNC also called principle of explosion sometimes.

cinch
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  • Just so I understand, what in my comment suggests the conflation of the two concepts? By establishing that "Shaun" concluded his argument with $\varphi(a) \land \neg \varphi(a)$, this was evidence of a contradiction with LNC (as the LNC concept would assert $\neg \big (\varphi(a) \land \neg \varphi(a) \big)$. – S.C. Aug 17 '21 at 23:31
  • @S.Cramer sorry for any misunderstanding if any. Under classic logic system in most textbooks, you're correct to say "this is making use of the principle of non-contradiction...therefore your final claim reads as φ(a)∧¬φ(a)." No issue here since essentially Shaun's method is to prove LEM via LNC. What I want to emphasize is that LEM and LNC are 2 independent equally valid principles here, but not always, in BHK intuitionistic logic. Here is a post I found useful to further clarify. – cinch Aug 18 '21 at 00:22
  • @S.Cramer after reading my linked references, actually as a further question (as you seem to try to learn more meaningful things), in BHK intuitionistic system we know your original LEM like question/claim no longer holds, while Shaun's above method direction can still apply even in BHK where LNC still holds, where's the issue then in the same steps will fail to prove your original LEM? – cinch Aug 18 '21 at 00:34
  • Are you asking if Shaun's proof "works" in an intuitionistic framework? If so, my hunch is no because, in the absence of the double negation axiom, $\neg \left (\varphi(A) \lor \neg \varphi(A) \right)$ entails $\neg\varphi(A) \land \neg \neg \varphi (A)$. It is (at least to me) uncertain what $\neg \neg \varphi (A)$ reduces to. In the absence of that information, we cannot claim "$\neg \varphi(A) \land \varphi(A)$" and thus no contradiction with respect to LNC arises. Is that what you were probing for? (Thanks for the input) – S.C. Aug 18 '21 at 02:33
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    @S.Cramer sounds you fully got it right and I may over-concerned. Normally in BHK we can semantically interpret classic truth as constructible or provable. This sentence is unprovable is a true but unprovable case in BHK sense, unlike the famous liar paradox in classic logic. And double negation can be interpreted as The unprovability of an unprovable proposition is not the same as this proposition must be necessarily true or provable. – cinch Aug 18 '21 at 02:44