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?


closed as unclear what you're asking by D.W., FrankW, Wandering Logic, David Richerby, Yuval Filmus Apr 29 '14 at 23:35

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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