The settingAn argument from a set of formulas $\Delta$ is a pair $\langle \Phi, \alpha \rangle$ such that$\Phi \subseteq \Delta$$\Phi \nvdash \bot$$\Phi \vdash \alpha$ (this is what I am wondering about)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$$\langle \Phi, \alpha \rangle$ undercuts $\langle \Psi, \beta \rangle$ iff $\alpha \vdash \neg\beta \wedge \Psi$The exerciseThe original statement is $$\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))$$  My attemptThe left side is$$\{\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?
Propositional logic of arguments and undercuts
logic;propositional logic