Using existential generalization

[Terrell]

Homework Statement

is my method valid?

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

Homework Equations

N/A

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)

 

[fresh_42]

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.

 

[Stephen Tashi]

∃x¬R(x,a) --> ¬∃R(a,x)

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

Your general approach is correct.

 

[Terrell]

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

Your general approach is correct.

thank you stephen!

 

[Terrell]

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.

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

 

[fresh_42]

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

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.