I have a question where I was given the following atomic propositions:

Let H(x) = x can ski

Let P(x) = x plays soccer

Note: The universe of discourse is all humans

I was tasked to translate the following sentence logically:

No one who can ski plays soccer

I came up with two solutions for this sentence and I'm unsure if one is considered more correct:

  1. ∀x(¬(P(x)∧H(x))

  2. x (H(x) -> ~p(x))

Symbols Reference

  • $\begingroup$ How about $\forall x\in H(x),x\notin P(X)$? (Actually I don't know predicate logic but I am just guessing about this, is it right?) $\endgroup$ Commented Jan 25, 2018 at 2:08
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    $\begingroup$ Number one means "It's not the case that everyone can both ski and play soccer." Number two means "It's not the case that if a person can ski, they can't play soccer." Neither is really what you're aiming for. $\endgroup$ Commented Jan 25, 2018 at 2:11
  • $\begingroup$ @GaurangTandon: your formulas are not well formed. $H(x)$ is a sentence not a set. $\endgroup$ Commented Jan 25, 2018 at 2:11
  • $\begingroup$ @CheerfulParsnip But doesn't the sentence denote the set of all $x$ who can ski? $\endgroup$ Commented Jan 25, 2018 at 2:13
  • $\begingroup$ @GaurangTandon: No a sentence is a statement. You can of course form the set of all elements that satisfy the statement, but they are two different objects. $\endgroup$ Commented Jan 25, 2018 at 2:16

2 Answers 2


Your second sentence can be rescued by removing the negation at the beginning: $$\forall x (H(x)\to\neg P(x)).$$ This translates to, For every person, if they ski, they don't play soccer. This is equivalent in English to saying that nobody who skis plays soccer.

  • $\begingroup$ Is this second edited sentence considered to be more accurate of a translation than my first? I was aiming for a translation that wouldn't be as wordy with the if/then, thus keeping the structural integrity (and of course truth values) from english to logic $\endgroup$ Commented Jan 25, 2018 at 2:19
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    $\begingroup$ @JessicaTiberio You need to edit the first sentence too. It would be $\forall x\,(\neg(P(x)\wedge H(x))$. The edited sentences are logically equivalent. The second one seems closer to the English but that's subjective. $\endgroup$ Commented Jan 25, 2018 at 2:57
  • $\begingroup$ why that edit? What is the english differences between the two sentences? $\endgroup$ Commented Jan 25, 2018 at 3:11
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    $\begingroup$ Your original sentence means "It's not the case that everybody both skis and play soccer." This is equivalent to saying there is at least one person who doesn't ski and play soccer. That's different from saying that nobody does both. $\endgroup$ Commented Jan 25, 2018 at 3:17
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    $\begingroup$ ah makes sense now. I was attempting to negate the universal qualifier to change from "everyone" to "no one', not realizing that the negation of a universal qualifier is an existential qualifier and I was instead negating the wrong portion of my proposition. Makes a little more sense now! $\endgroup$ Commented Jan 25, 2018 at 3:27

According to my professor the following are equivalent ways to express no one who can ski plays soccer in logic.

¬∃xH(x) ∧ P(x)

≡ ∀x¬(H(x) ∧ P(x)) De Morgan's Law

≡ ∀x(¬H(x) ∨ ¬P(x)) De Morgan's Law

≡ ∀x(H(x) → ¬P(x)) Implication Relation

≡ ∀x(P(x) → ¬H(x)) Contrapositive

Thus, both my proposed options in the current version of this question are correct!


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