All girls are liked by some boys. No boy likes a sulk. No girl is a sulk.

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?

4 Answers

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)

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.

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:

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

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

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

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

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

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

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

and

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