# Trivial question about nested quantifiers.

Reading my textbook, I came across exercises for nested quantifiers.

The question: Let $L(x, y)$ be the statement “$x$ loves $y$,” where the domain for both $x$ and $y$ consists of all people in the world. Use quantiﬁers to express each of these statements.

i) Everyone loves himself or herself.

Textbook answer: $$\forall xL(x, x)$$

Is this equivalent to my answer? : $$\forall x\forall y((x=y)\to L(x,y))$$

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How do you know that the question is trivial? A good rule of the thumb with the word trivial is that one should only use it on things one can do... otherwise, it is a bit silly. –  Mariano Suárez-Alvarez Feb 23 '13 at 19:29
I am sorry, I was using it in a self-deprecating manner. I am not so sure if my questions are dumb or not. –  Chase Feb 23 '13 at 19:32
@Chase Why don't you try to prove that both answers are equivalent? –  Git Gud Feb 23 '13 at 19:33
Don't be self-deprecating then. –  Mariano Suárez-Alvarez Feb 23 '13 at 19:40

They're equivalent.

Let's start with $(\forall x)(\forall y) ((x=y)\to L(x,y))$:

1. Apply instantiation twice and set $x/x$ and $x/y$, getting $(x=x)\to L(x,x)$.
2. $x=x$ is an axiom of reflexivity of equality.
3. From these two infer $L(x,x)$ (modus ponens).
4. Apply generalization to get $(\forall x) L(x,x)$.

And the other way around: Let's start with $(\forall x) L(x,x)$.

1. Apply instantiation to get $L(x,x)$.
2. Use substitution for formulas axiom to get $(x = y) \to (L(x,x) \to L(x,y))$.
3. Exchange the antecendents to get $L(x,x) \to ((x = y) \to L(x,y))$ (we can do this because $\phi\to(\psi\to\rho) \equiv \psi\to(\phi\to\rho)$ is a tautology).
4. Apply modus ponens on 3. and 1. to get $(x = y) \to L(x,y)$.
5. Use generalization twice to get $(\forall x)(\forall y) (x = y) \to L(x,y)$
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