first-order logic translation

Question

I have an argument that I am trying to translate into first order logic. I understand that there's meant to be a symbol '=' in the answer but I'm unsure where.

"Some things are neither rough nor ugly, but anything that is rough is ugly. So my friend dog is both rough and ugly, because anything that’s not rough isn’t my friend dog."

This is what I've done so far:

$\exists$ = there exists, $\forall$ = for all, $\land$ = and, $\lor$ = or, $R$ = rough, $U$ = ugly, $D$ = dog, $\lnot$ = not

$\exists x~(\lnot Rx \land \lnot Ux), \forall x~(Rx \to Ux), \forall x~(\lnot Rx \to \lnot Dx) \vdash \exists x~(Dx\color{red}\land Rx \land Ux)$

So there must be an = somewhere, I'm thinking it will be in the conclusion.

Answer 1

I understand that there's meant to be a symbol '=' in the answer but I'm unsure where.

"Some things are neither rough nor ugly, but anything that is rough is ugly. So my friend dog is both rough and ugly, because anything that’s not rough isn’t my friend dog."

The equality claim is clearly in the "anything that’s not rough isn’t my friend dog," clause; to be precise, it is a negation of an equality.

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$\exists$ = there exists, $\forall$ = for all, $\land$ = and, $\lor$ = or, $R$ = rough, $U$ = ugly, $D$ = dog, $\lnot$ = not

$\exists x~(\lnot Rx \land \lnot Ux), \forall x~(Rx \to Ux), \forall x~(\lnot Rx \to \lnot Dx) \vdash \exists x~(Dx \color{red}\land Rx \land Ux)$

• "Something is not rough and not ugly" checks okay $(\checkmark)$.   So too would $\exists x~\neg(Rx\lor Ux)$.
• "Any rough thing is ugly" also $(\checkmark)$.
• "Any thing not rough is not my friend dog" too $(\checkmark)$.
• Finally I presumed that you had dropped the conjunction because: "Some thing is my friend dog, it is rough, and it is ugly" would check out okay as well.

So there must be an = somewhere, I'm thinking it will be in the conclusion.

You are using $D$ as a predicate, standing for "is my friend dog", rather than using $d$ as a constant standing for "my friend dog".   That is where the equality claim lives.

• So $Dx$ means $(x=d)$.

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So you indeed have a translation of the explicit argument given. However, it is missing the implicit assumption that there is such a thing called "my friend dog".

$$\exists x~(\lnot Rx \land \lnot Ux), \forall x~(Rx \to Ux), \forall x~(\lnot Rx \to \lnot Dx)\color{blue}{, \exists x~Dx} \vdash \exists x~(Dx \land Rx \land Ux)$$