What does the disjunction elimination rule say? - MathOverflow

What does the disjunction elimination rule say?

2010-02-26T09:55:17Z

I read about two different versions of the disjunction elimination rule.

The first version (http://www.fecundity.com/logic/) says that:

if $\Sigma\vdash\phi_0\lor\phi_1$ and $\Sigma\vdash\lnot\phi_0$, then $\Sigma\vdash\phi_1$
if $\Sigma\vdash\phi_0\lor\phi_1$ and $\Sigma\vdash\lnot\phi_1$, then $\Sigma\vdash\phi_0$

The second version (S. Hedman - A First Course in Logic) says that:

if $\Sigma\models\phi_0\lor\phi_1$, $\Sigma\cup\left\{\phi_0\right\}\vdash\phi_2$ and $\Sigma\cup\left\{\phi_1\right\}\vdash\phi_2$, then $\Sigma\vdash\phi_2$

Using the first version of the rule, I can't even demonstrate that if $\Sigma\vdash\phi\lor\phi$ then $\Sigma\vdash\phi$. Perhaps the entire system presented by the first source is not complete, in the sense that you can't prove certain true statements. Of course, it may be my fault, instead.

Thanks.

Answer by Neel Krishnaswami for What does the disjunction elimination rule say?

2010-02-26T10:09:04Z

Yes, the first version of your rule is incomplete. In natural deduction style, the elimination rule for disjunctions is:

$$\frac{\Gamma \vdash A \vee B \qquad \Gamma, A\vdash C \qquad \Gamma, B\vdash C}{\Gamma \vdash C}$$

This corresponds to the second version of your rule.

Answer by Charles Stewart for What does the disjunction elimination rule say?

2010-02-26T16:45:40Z

The first rule is not the regular disjunction elimination rule, but is known as disjunctive syllogism, and is essentially the modus tollendo ponens rule of term logic. The two rules are mutually admissible in reasonable formulations of classical logic, but the first rule is strictly weaker in intuitionistic logic.