$r:$All prime numbers are either even or odd, Is it a true statement?

I was studying Mathematical Logic then i came across above question. Since here connecting word is "OR" so if i separate two statement then it become

$p:$ All the prime number are even

$q:$ All the prime number are odd

Because both statement $p$ and $q$ are false so final statement $r$ must be false using truth value of statement for "OR" connective.

But my intuition says $r$ is true.

Am i thinking correct?

Please Help me in this.

  • 1
    $\begingroup$ You may be hearing it incorrectly -- what I first thought the title of the question meant. If the statement were "all numbers [integers] are even or odd", generally we'd say, sure that's true, as every integer $n$ is either even or odd. But that's not what's meant: it's "either all primes or even or all primes are odd", not the true statement "for all primes $p$, either $p$ is even or $p$ is odd". $\endgroup$
    – BrianO
    Jan 7, 2022 at 5:27
  • 2
    $\begingroup$ The natural language statement you have to work with is ambiguous — at least, it is for you! Recall or notice that $\forall x\,(A \lor B)$ does not imply, but is implied by, $\forall x\,A \lor \forall x\,B$. $\endgroup$
    – BrianO
    Jan 7, 2022 at 5:36
  • 1
    $\begingroup$ You are using ambiguity of language. "(All prime numbers are even) OR (All prime numbers are odd)" is certainly FALSE. $2$ is a prime number that is even but not odd. And $3$ is a prime number that is odd but not even. But "All prime numbers are (even or odd)" is certainly TRUE. Every prime number is an integer. And every integer is itself either even or it is odd. But "(All X are a) OR (All x are b)" is a VERY different statement than "All X are (a or b)". $\endgroup$
    – fleablood
    Jan 7, 2022 at 19:31

2 Answers 2


You erroneously distributed the “all”. The correct interpretation is:

For every prime $p$: $p$ is even or $p$ is odd

Since every integer is either even or odd, we have:

($p$ is even or $p$ is odd) is true for all primes $p$

Therefore it is true — as your intuition suggested


$✔\quad$ All numbers are either even or odd.

$✗\quad$ Either all numbers are even, or all numbers are odd.

In general, $$∀x \;\Big(A(x)\text{ or }B(x)\Big)\quad\text{does not imply}\quad∀x A(x)\;\text{ or }\;∀x B(x);$$ however, $$∃x \;\Big(A(x)\text{ or }B(x)\Big)\quad\text{is equivalent to}\quad∃x A(x)\;\text{ or }\;∃x B(x),$$ and $$∀x \;\Big(A(x)\text{ and }B(x)\Big)\quad\text{is equivalent to}\quad∀x A(x)\;\text{ and }\;∀x B(x).$$

  • $\begingroup$ For completeness' sake: $$∃x A(x)\;\text{ and }\;∃x B(x)\quad\text{does not imply}\quad∃x \;\Big(A(x)\text{ and }B(x)\Big).$$ $\endgroup$
    – ryang
    Jan 8, 2022 at 4:37

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