In the book "How to Prove It", by Velleman, appear these two examples.

  1. Prove that for every integer $x$, the remainder when $x^2$ is divided by 4 is either 0 or 1.
  2. Prove that for every real number x, if $x^2 \geq x$ then either $x \leq 0 \lor x \geq 1$.


  1. $x \in \mathbb{Z} \to (x^2 \text{ has remainder } 0) \lor (x^2 \text{ has remainder } 1)$
  2. $\forall x(x^2 \geq x \to (x \leq 0 \lor x \geq 1)$

In both cases, the author symbolises the conclusion with a disjunction.

In my mind, the conclusion is an exclusive or. In the first example, if I write a number in the form $k \in \mathbb{Z}$, that same number can not be written in the form $4l$ for some number $l \in \mathbb{Z}$. In the second one, a number being 0 excludes the possiblity of it being greater than or equal to 1.

What would be the explanation for this issue, from a logic perspective ?

If my perspective is incorrect, how does a proof of an "exclusive or" look like ?

  • $\begingroup$ I think this covers the main aspects of the question. $\endgroup$ – Gae. S. Apr 25 '20 at 21:17
  • $\begingroup$ Why in the symbolization of 1. appears $\mathbb{R}$ instead of $\mathbb{Z}$? $\endgroup$ – Riccardo Apr 25 '20 at 21:35

Since in these cases the options are mutually exclusive, both kinds of or are equivalent. It's usually more convenient in mathematics to use $\lor$ if possible, because it's dual to $\land$, and many proofs easily obtain a result of the form $p\lor q$, where $p\oplus q$ is either harder to prove or in general false. For example, $xy=0\to x=0\lor y=0$ (unless we have zero divisors).

  • $\begingroup$ Could you expand a little bit on your example $xy=0\to x=0\lor y=0$ ? $\endgroup$ – F. Zer Apr 25 '20 at 22:22
  • 1
    $\begingroup$ @F.Zer The product of nonzero quantities is nonzero, so from $xy=0$ we can deduce $x=0\lor y=0$. But that's all we can deduce; we can't deduce $x=0\oplus y=0$. $\endgroup$ – J.G. Apr 25 '20 at 22:24

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