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Theorem hbbi 1456
Description: If 𝑥 is not free in 𝜑 and 𝜓, it is not free in (𝜑𝜓). (Contributed by NM, 5-Aug-1993.)
Hypotheses
Ref Expression
hb.1 (𝜑 → ∀𝑥𝜑)
hb.2 (𝜓 → ∀𝑥𝜓)
Assertion
Ref Expression
hbbi ((𝜑𝜓) → ∀𝑥(𝜑𝜓))

Proof of Theorem hbbi
StepHypRef Expression
1 dfbi2 374 . 2 ((𝜑𝜓) ↔ ((𝜑𝜓) ∧ (𝜓𝜑)))
2 hb.1 . . . 4 (𝜑 → ∀𝑥𝜑)
3 hb.2 . . . 4 (𝜓 → ∀𝑥𝜓)
42, 3hbim 1453 . . 3 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
53, 2hbim 1453 . . 3 ((𝜓𝜑) → ∀𝑥(𝜓𝜑))
64, 5hban 1455 . 2 (((𝜑𝜓) ∧ (𝜓𝜑)) → ∀𝑥((𝜑𝜓) ∧ (𝜓𝜑)))
71, 6hbxfrbi 1377 1 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 101  wb 102  wal 1257
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 103  ax-ia2 104  ax-ia3 105  ax-5 1352  ax-gen 1354  ax-4 1416  ax-i5r 1444
This theorem depends on definitions:  df-bi 114
This theorem is referenced by:  euf  1921  sb8euh  1939
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