# Using existential generalization

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1. Feb 10, 2017

### Terrell

1. The problem statement, all variables and given/known data
is my method valid?

∃x¬R(x,a) --> ¬∃R(a,x)
¬R(a,a)
thus, ¬R(a,b)

2. Relevant equations
N/A

3. The attempt at a solution
∃x¬R(x,a) by existential gen. of ¬R(a,a)
¬∃R(a,x) by modus ponens
∀x¬R(a,x) by identity of ¬∃R(a,x)
¬R(a,b) by universal instantiation of ∀x¬R(a,x)

2. Feb 10, 2017

### Staff: Mentor

I have trouble to understand your notation. $\lnot R(x,a)$ suggests, that it is a statement, whereas $\exists R(a,x)$ looks like an element somewhere. Furthermore $a$ isn't quantified.

3. Feb 10, 2017

### Stephen Tashi

As fresh_42 pointed out, the usual notation would be something like $\lnot( \exists x ( R(a,x) )$

4. Feb 10, 2017

### Terrell

thank you stephen!

5. Feb 10, 2017

### Terrell

yes i missed an "x" there. how was "a" not quantified?

6. Feb 10, 2017

### Staff: Mentor

Well, is $a$ fixed? Then it could be taken as part of $R$, if $R$ is symmetric, which I don't know. Or does it mean for all $a$, or there is an $a$? Since it is in all occurrences of $R$ I tend to put it into the definition of $R$ to get rid of it, as it seems to be unnecessary. But then there are $R(a,x)$ and $R(x,a)$ and I don't know. what is the difference between them.