Metamath Proof Explorer


Theorem hb3an

Description: If x is not free in ph , ps , and ch , it is not free in ( ph /\ ps /\ ch ) . (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)

Ref Expression
Hypotheses hb.1
|- ( ph -> A. x ph )
hb.2
|- ( ps -> A. x ps )
hb.3
|- ( ch -> A. x ch )
Assertion hb3an
|- ( ( ph /\ ps /\ ch ) -> A. x ( ph /\ ps /\ ch ) )

Proof

Step Hyp Ref Expression
1 hb.1
 |-  ( ph -> A. x ph )
2 hb.2
 |-  ( ps -> A. x ps )
3 hb.3
 |-  ( ch -> A. x ch )
4 df-3an
 |-  ( ( ph /\ ps /\ ch ) <-> ( ( ph /\ ps ) /\ ch ) )
5 1 2 hban
 |-  ( ( ph /\ ps ) -> A. x ( ph /\ ps ) )
6 5 3 hban
 |-  ( ( ( ph /\ ps ) /\ ch ) -> A. x ( ( ph /\ ps ) /\ ch ) )
7 4 6 hbxfrbi
 |-  ( ( ph /\ ps /\ ch ) -> A. x ( ph /\ ps /\ ch ) )