Proposition 41: Generalized conjunction introduction (inference rule)

Given that the statements \alpha, \beta, ..., \omega hold, we may infer that the generalized conjunction (\alpha \wedge \beta \wedge \cdots \wedge \omega) holds: \cfrac{\vdash \alpha \quad \vdash \beta \quad \cdots \quad \vdash \omega}{\vdash (\alpha \wedge \beta \wedge \cdots \wedge \omega)}.

Proof: We argue using structural induction over the recursive definition of generalized conjunctions.

Base cases: Suppose that the statements \alpha , \beta and \gamma hold. Using conjunction introduction, the conjunction (\alpha \wedge \beta) holds. By the same rule, \big( (\alpha \wedge \beta) \wedge \gamma \big) then holds. But (\alpha \wedge \beta \wedge \gamma) is short for precisely this statement.

Inductive step: We take as the induction hypothesis that the desired result is evident for statements \alpha, \beta, ..., \psi , and want to show from this hypothesis that it is evident for statements \alpha, \beta, ..., \psi, \omega . But if all of these statements hold then our hypothesis gives us that (\alpha \wedge \beta \wedge \cdots \wedge \psi) holds; conjunction introduction then lets us infer that \big( (\alpha \wedge \beta \wedge \cdots \wedge \psi) \wedge \omega) \big) holds. But this statement is abbreviated by (\alpha \wedge \beta \wedge \cdots \wedge \psi \wedge \omega) . \square