For all

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If [math]P [/math] is a predicate that depends on [math]x [/math], then "for all [math]x [/math], [math]P [/math]" is a proposition. It is true if every possible value of [math]x [/math] makes [math]P [/math] evaluate to true.

  • If your goal is to prove "for all [math]x\href{/cs2800/wiki/index.php/%E2%88%88}{∈}A [/math], P", you can proceed by choosing an arbitrary value [math]x∈A [/math] and then proving that P holds for that [math]x [/math].

The fact that [math]x [/math] is arbitrary does not mean you get to pick [math]x [/math]; on the contrary, your proof should work no matter what [math]x [/math] you choose. This means you can't use any property of [math]x [/math] other than that [math]x \href{/cs2800/wiki/index.php/%E2%88%88}{∈} A [/math].

  • If you know [math]P [/math] holds for all [math]x [/math], then you can conclude [math]P [/math] holds for any specific [math]x [/math]. For example, if you know for all [math]x \href{/cs2800/wiki/index.php/%E2%88%88}{∈} ℝ [/math], [math]x^2 ≥ 0 [/math], then you can conclude [math]7^2 ≥ 0 [/math] (since [math]7 ∈ ℝ [/math]).