3
$\begingroup$
  1. All girls are liked by some boys.
  2. No boy likes a sulk.
  3. No girl is a sulk.

How do you write the above in predicate logic statements without starting with a negation? My guess is below. Let me know if you differ.

  1. ∀x ∃y girl(x) → boy(y) ∧ like(y,x)
  2. ∀x ∃y boy(x) → sulk(y) ∧ ~like(x,y)
  3. ∀x girl(x) → ~ sulk(y)

After that, how do you prove it?

$\endgroup$
  • $\begingroup$ 2) as you wrote it is. For every boy there is a sulk that the boy doesn't like. That's not what the statement means. Why don't you want a negative? That is what statement two is: There isn't any sulk that is liked by any boy (or for every boy there doesn't exist any sulk that is liked by the boy). Why don't you want a negative? $\endgroup$ – fleablood Sep 15 at 5:21
  • $\begingroup$ 3) is incorrect (as well as 2, which has already been mentioned), it should only involve $x$ and not $y$ $\endgroup$ – Morgan Rodgers Sep 15 at 5:22
  • $\begingroup$ Can you do a dual subject? For all boys, x and all sulks y, then x does not like y. $\endgroup$ – fleablood Sep 15 at 5:23
3
$\begingroup$

How do you write the above in predicate logic statements without starting with a negation?

First write them starting with a negation, then apply DeMorgan's Laws to "move it inwards". Also use Implication Equivalence when applicable. $${\text{No boy likes a sulk}\\\neg\exists x~\exists y~(B(x)\wedge S(y)\wedge L(x,y))\\\forall x~\neg\exists y~(B(x)\wedge S(y)\wedge L(x,y))\\\forall x~\forall y~\neg(B(x)\wedge S(y)\wedge L(x,y))\\\forall x~\forall y~(\neg(B(x)\wedge S(y))\vee\neg L(x,y))\\\forall x~\forall y~(B(x)\wedge S(y)\to\neg L(x,y))}$$

$\forall x~\forall y~(B(x)\to(S(y)\to\neg L(x,y)))$ is also acceptable. (Re: the exportation identity)

$\endgroup$
2
$\begingroup$

I would do it the following way:

$1. \forall x \exists y \exists z(G(x) \to ((y \neq z) \land L(yx) \land L(zx)) $, because it seems that there's more than one boy per girl satisfying that the former likes the latter.

$2. \forall x \forall y ((B(x) \land S(y)) \to \neg L(xy))$, because given a boy and a ulk, the boy wouldn't like the sulk (it doesn't matter if the sulk is a girl, because I'm progressive politically).

$3. \forall x (G(x) \to \neg S(x))$

Here $G$, $B$ and $S$ are the girl, boy and sulk predicates, respectively, and $L$ is the relation meaning that the first likes the second.

$\endgroup$
1
$\begingroup$

What you have put here is almost all correct. Make sure you read up on the rules of logic so that you know what is going on here.

  1. All girls are liked by some boys. This is equivalent to:

$$\forall x ,G_x \implies \exists y \ni(B_y \wedge L_{(x,y)})$$

  1. No boy likes a sulk. This is equivalent to:

$$\neg \exists x,y\ni B_x \wedge S_y\wedge L_{(x,y)}$$ $$\forall x,y,\neg( B_x \wedge S_y\wedge L_{(x,y)})$$ $$\forall x,y,\neg B_x \vee \neg S_y \vee \neg L_{(x,y)})$$ $$\forall x,y, B_x \implies (\neg S_y \vee \neg L_{(x,y)})$$ $$\forall x,y, B_x \implies (S_y \implies \neg L_{(x,y)})$$ $$\forall x,y, (B_x \wedge S_y) \implies \neg L_{(x,y)}$$

  1. No girl is a sulk. This is equivalent to:

$$\neg \exists x \ni G_x \wedge S_x$$ $$\forall x, \neg (G_x \wedge S_x)$$ $$\forall x, \neg G_x \vee \neg S_x$$ $$\forall x, G_x \implies \neg S_x$$

$\endgroup$
  • $\begingroup$ The symbol $\ni$ is from set theory and does not make sense here. $\endgroup$ – miracle173 Sep 15 at 7:28
  • $\begingroup$ @miracle173 - notations abound. Please don't assume that if you see one you don't recognize, it is automatically wrong. It is evident that this one means "such that". $\endgroup$ – Paul Sinclair Sep 15 at 14:58
1
$\begingroup$

I think 2. and 3. should be of the form

  1. $\forall x \forall y \;\text{boy}(x)\land\text{sulk} (y) \rightarrow \lnot \;\text{like}(x,y)$

and

  1. $\forall x\; \text{girl}(x) \rightarrow \lnot\;\text{sulk}(x)$
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.