If \( x \) is not free in \( \varphi \), \( \psi \), and \( \chi \), it is not free in
\( \varphi \wedge \psi \wedge \chi \). (Contributed by NM, 14-Sep-2003.) (Proof
shortened by Wolf Lammen, 2-Jan-2018.) Reduce axiom dependencies.
(Revised by SN and Sky Winter, 23-Oct-2026.)
\( \vdash \varphi \wedge \psi \wedge \chi \rightarrow \forall x \, (\varphi \wedge \psi \wedge \chi) \)
\( \vdash \varphi \wedge \psi \wedge \chi \leftrightarrow \varphi \wedge \psi \wedge \chi \)
\( \vdash \varphi \wedge \psi \wedge \chi \rightarrow \forall x \, (\varphi \wedge \psi \wedge \chi) \)