Theorem hba1

The setvar \( x \) is not free in \( \forall x \, \varphi \). This corresponds to the axiom (4) of modal logic. Example in Appendix in [Megill] p. 450 (p. 19 of the preprint). Also Lemma 22 of [Monk2] p. 114. (Contributed by NM, 24-Jan-1993.) (Proof shortened by Wolf Lammen, 12-Oct-2021.)

Conclusion
\( \vdash \forall x \, \varphi \rightarrow \forall x \, \forall x \, \varphi \)
Proof
nfa1
nf5ri _