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.



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.

  • $\begingroup$ That's a good point, Charles -- I am sorry to say I forgot entirely about classical logic! $\endgroup$ – Neel Krishnaswami Feb 26 '10 at 17:40

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.


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