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Theorem hban 1436
Description: If x is not free in φ and ψ, it is not free in (φ ψ). (Contributed by NM, 5-Aug-1993.) (Proof shortened by Mario Carneiro, 2-Feb-2015.)
Hypotheses
Ref Expression
hb.1 (φxφ)
hb.2 (ψxψ)
Assertion
Ref Expression
hban ((φ ψ) → x(φ ψ))

Proof of Theorem hban
StepHypRef Expression
1 hb.1 . . 3 (φxφ)
2 hb.2 . . 3 (ψxψ)
31, 2anim12i 321 . 2 ((φ ψ) → (xφ xψ))
4 19.26 1367 . 2 (x(φ ψ) ↔ (xφ xψ))
53, 4sylibr 137 1 ((φ ψ) → x(φ ψ))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wal 1240
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-5 1333  ax-gen 1335
This theorem depends on definitions:  df-bi 110
This theorem is referenced by:  hbbi  1437  hb3an  1439  hbsbv  1814  mopick  1975  eupicka  1977  mopick2  1980  cleqh  2134
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