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The setting

An argument from a set of formulas $\Delta$ is a pair $\langle \Phi, \alpha \rangle$ such that

  1. $\Phi \subseteq \Delta$
  2. $\Phi \nvdash \bot$
  3. $\Phi \vdash \alpha$ (this is what I am wondering about)
  4. There is no $\Phi' \subset \Phi$ such that $\Phi' \vdash \alpha$.

I am trying to think of how this undercut works.

If $\langle \Phi, \alpha \rangle$ and $\langle \Phi, \beta \rangle$ are arguments,then:

  • $\langle \Phi, \alpha \rangle$ rebuts $\langle \Phi, \beta \rangle$ iff $\alpha \vdash \neg\beta$
  • $\langle \Phi, \alpha \rangle$ undercuts $\langle \Psi, \beta \rangle$ iff $\alpha \vdash \neg\beta \wedge \Psi$

The exercise

The original statement is $$\langle\{\gamma,\gamma\rightarrow\neg\beta\},\neg(\beta\wedge(\beta\rightarrow\alpha))\rangle \quad \text{undercuts} \quad \langle\{\beta,\beta\rightarrow\alpha\},\alpha\rangle$$

I believe these two should both be arguments.

$$\{\gamma,\gamma\rightarrow\neg\beta\} \vdash \neg(\beta\wedge(\beta\rightarrow\alpha))$$

My attempt

The left side is

$$\{\gamma,\gamma\rightarrow\neg\beta \} \vdash \neg\beta$$

The right side is

$$\neg(\beta\wedge(\beta\rightarrow\alpha)) \vdash \neg(\beta\wedge\neg\beta\vee\alpha) \vdash \neg\alpha$$

Is my proof right? What is the next step?

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  • $\begingroup$ In order to produce a formal proof, we need to know what axioms you're using. Assuming we have modus ponens, you're correct in deducing $\neg \beta$. Assuming you have access to "(X) proves (X or Y)" and De Morgan's laws, this is fairly straightforward. $\endgroup$ – Patrick87 Apr 25 '14 at 16:36
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    $\begingroup$ The right side does not imply $\neg \alpha$, you're missing some parenthesis: $\beta \wedge \left(\beta \rightarrow \alpha\right) = \beta \wedge \left(\neg \beta \vee \alpha\right)$, so the negation is actually $\neg \beta \vee \neg \left(\neg \beta \vee \alpha\right) = \neg \beta \vee \left(\beta \wedge \neg \alpha\right) = \neg \beta \vee \neg \alpha = \beta \rightarrow \neg \alpha$. $\endgroup$ – Jared Apr 26 '14 at 6:03
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    $\begingroup$ And if you take that, along with the left side, that $\gamma$ is given, therefore (by modus ponens) $\neg \beta$ is true, this implies nothing about $\alpha$ (it could be true or false). $\endgroup$ – Jared Apr 26 '14 at 6:10