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Theorem hbbi 1413
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 (φxφ)
hb.2 (ψxψ)
Assertion
Ref Expression
hbbi ((φψ) → x(φψ))

Proof of Theorem hbbi
StepHypRef Expression
1 dfbi2 368 . 2 ((φψ) ↔ ((φψ) (ψφ)))
2 hb.1 . . . 4 (φxφ)
3 hb.2 . . . 4 (ψxψ)
42, 3hbim 1410 . . 3 ((φψ) → x(φψ))
53, 2hbim 1410 . . 3 ((ψφ) → x(ψφ))
64, 5hban 1412 . 2 (((φψ) (ψφ)) → x((φψ) (ψφ)))
71, 6hbxfrbi 1334 1 ((φψ) → x(φψ))
Colors of variables: wff set class
Syntax hints:  wi 4   wa 97  wb 98  wal 1221
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 1309  ax-gen 1311  ax-4 1373  ax-i5r 1401
This theorem depends on definitions:  df-bi 110
This theorem is referenced by:  euf  1878  sb8euh  1896
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