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Theorem hbbi 1483
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 380 . 2 ((𝜑𝜓) ↔ ((𝜑𝜓) ∧ (𝜓𝜑)))
2 hb.1 . . . 4 (𝜑 → ∀𝑥𝜑)
3 hb.2 . . . 4 (𝜓 → ∀𝑥𝜓)
42, 3hbim 1480 . . 3 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
53, 2hbim 1480 . . 3 ((𝜓𝜑) → ∀𝑥(𝜓𝜑))
64, 5hban 1482 . 2 (((𝜑𝜓) ∧ (𝜓𝜑)) → ∀𝑥((𝜑𝜓) ∧ (𝜓𝜑)))
71, 6hbxfrbi 1404 1 ((𝜑𝜓) → ∀𝑥(𝜑𝜓))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 102  wb 103  wal 1285
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1379  ax-gen 1381  ax-4 1443  ax-i5r 1471
This theorem depends on definitions:  df-bi 115
This theorem is referenced by:  euf  1950  sb8euh  1968
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