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Let A(x) and B(x) be predicates over the same nonempty domain. State True/False for each statement. If True give non formal/formal proof in English. If False give a counter example.

1) ∃x, (∼ A(x)) equivalent to ∃x, (A(x) → B(x))

2) ∀x, (A(x) → B(x)) logically implies (∀x, A(x)) → (∀x, B(x))

3) ∃x ∀y, (A(x) → B(y)) equivalent to (∃x, ∼ A(x)) ∨ (∀y, B(y))

So far I figured that 1 is False by if-then as or. 2 most likely true by math distributive properties but I cant prove it. and 3 is True by If-then as or. I need help writing formal or non formal proof to show that my statements are correct.

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    Any formal proof needs a formal system, i.e. a proof system, like Natural Deduction or Tableau or... – 2017-02-05
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    To show that 1) is false, you have to find a suitable counterexample; consider as domain $\mathbb N$ and $x \ge 0$ as $A(x)$ and $x > 0$ as $B(x)$. – 2017-02-05

1 Answers 1

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To show any of these False, all you need is a counterexample, which specifies what kinds of things you quantify over (the domain), and what the predicates mean.

  1. is indeed False. For a counterexample, you could use as the domain all natural numbers, $A(x)$: $x$ is a number, and $B(x)$: $x$ is an even number. In that case, since there is no number that is not a number, $\forall x \neg A(x)$ is false. But, since for all even numbers it is true that if it is a number, then it is an even number, the statement $\exists x (A(x) \to B(x))$ is true. Therefore, the two statements are not equivalent.

  2. Is indeed True. OK, since you can just do a proof in English, let's show that if you assume $\forall x (A(x) \to B(x))$, then it will also be true that $\forall x A(x) \to \forall B(x)$. And to show the latter, assume that $\forall x A(x)$, and show that $\forall B(x)$. And finally, to show that $\forall B(x)$, you need to show that any arbitrary object has property $B$, given that $\forall x (A(x) \to B(x))$ and $\forall x A(x)$. OK, well, if you have any arbitrary object, call it $X$, then given $\forall x A(x)$ that means that $X$ has property $A$. But we also have $\forall x (A(x) \to B(x))$, which states that anything that has property $A$ will also have property $B$. So, $X$ has property $B$., But since $X$ was arbitrary, that means that everything has property $B$. Hence, $\forall B(x)$, as desired.

  3. Is indeed True ... but I'll leave the proof to you.

Hope this helps!