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Theorem falimd 1329
Description: implies anything. (Contributed by Mario Carneiro, 9-Feb-2017.)
Assertion
Ref Expression
falimd ((φ ⊥ ) → ψ)

Proof of Theorem falimd
StepHypRef Expression
1 falim 1328 . 2 ( ⊥ → ψ)
21adantl 452 1 ((φ ⊥ ) → ψ)
Colors of variables: wff setvar class
Syntax hints:  wi 4   wa 358  wfal 1317
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8
This theorem depends on definitions:  df-bi 177  df-an 360  df-tru 1319  df-fal 1320
This theorem is referenced by: (None)
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