Proposition 24: Provided a variable x occurs in at least one of the statements \alpha, \beta, ..., \omega , it also occurs in the iterated conjunction (\alpha \wedge \beta \wedge \cdots \wedge \omega).

Proof: We shall argue by means of structural induction over the definition of iterated conjunction.

Base cases: Suppose x occurs in at least one of \alpha , \beta and \gamma . Then, x occurs in at least one of \alpha or \beta , or else x occurs in \gamma . In the former case, x occurs in the conjunction (\alpha \wedge \beta) , so it occurs in \big( (\alpha \wedge \beta) \wedge \gamma) by (d) in the definition of occurrence. In the latter case, x occurs in this same statement by the same rule. But this statement is the definition of (\alpha \wedge \beta \wedge \gamma) .

Inductive step: We take as the induction hypothesis that the desired result is known for statements \alpha, \beta, ..., \psi , and want to show from this that it is also justified 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 \omega . 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