Proposition 14: If a variable x occurs in at least one of the statements \alpha, \beta, ..., \omega , then it occurs in the generalized conjunction (\alpha \wedge \beta \wedge \cdots \wedge \omega) .

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

Base cases: If x occurs in at least one of \alpha , \beta and \gamma , there are two possible cases: x occurs in at least one of \alpha or \beta , or x occurs in \gamma . In the former case, x occurs in (\alpha \wedge \beta) , so it occurs in \big( (\alpha \wedge \beta) \wedge \gamma) by (d) in the definition of occurrence. On the latter case, x occurs in this same statement by the same rule. But by this statement is abbreviated by (\alpha \wedge \beta \wedge \gamma) .

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 that it is also evident for \alpha, \beta, ..., \psi, \omega . If x occurs in at least one of \alpha, \beta, ..., \psi, \omega , then it either occurs in at least one of \alpha, \beta, ..., \psi or it occurs in \gamma . In the former case, x occurs in (\alpha \wedge \beta \wedge \cdots \wedge \psi) by the induction hypothesis. Thus, in either of the cases, it occurs inn \big( (\alpha \wedge \beta \wedge \cdots \wedge \psi) \wedge \omega\big) by (d) in the definition of occurrence. But (\alpha \wedge \beta \wedge \cdots \wedge \psi \wedge \omega) is short for just this statement. \square