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I'm trying to follow a proof in a logic text and it seems like the author used universal instantiation twice to reach the needed result. I was under the impression that you could only use UI one time in a proof.

user166057
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    Where did you get that impression? – Andreas Blass Jul 24 '14 at 04:26
  • This may not be particularly interesting to say, but it is indeed that case that, for all real numbers $a$ there exists a real number $c$ such that, for all positive numbers $e$, $e>a+c$. – Micah Jul 24 '14 at 04:27
  • I've only ever seen it used once in the proofs I've seen. Intuitively it seems like I should be able to use UI as many times as I want since the statement is true for all objects in the universe. Is this a correct assumption? – user166057 Jul 24 '14 at 04:28
  • If technique X can only be used once in a proof, and I use it to prove A implies B, and use it to prove B implies C, then I wouldn't be able to combine these proofs to show A implies C as that would require two uses, and mathematics would pretty well come to a halt. – Dan Piponi Jul 24 '14 at 04:39

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Nothing prevents us to use universal instantiation (UI) more than once in a proof.

Consider the formula in the language of formal arithmetics :

$\forall x \exists y (y = S(x))$

where $S(x)$ is the "successor" function.

We can clearly instantiate it with $0$, to get :

$\exists y (y = S(0))$

which is true when $y$ get the value $1$; and again with :

$\exists y (y = S(S(0)))$

which is true when $y$ get the value $2$.

Added

You can see Peter Smith's answer to this post.

In order to prove :

$\forall x \forall yR(x,y) \vdash \forall xR(x,x)$

we have to proceed this way :

(i) $\forall x \forall yR(x,y)$ --- assumed

(ii) $\forall yR(x,y)$ --- by Universal Instantiation

(iii) $R(x,x)$ --- by UI

(iv) $\forall x R(x,x)$ --- by Universal Generalization, $x$ is not free in $\Gamma$, where $\Gamma = \{ \forall x \forall yR(x,y) \}$.

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Mauro ALLEGRANZA
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