Theorem falimd 1329

Description: ⊥ implies anything. (Contributed by Mario Carneiro, 9-Feb-2017.)

Assertion

Ref	Expression
falimd	⊢ ((φ ∧ ⊥ ) → ψ)

Proof of Theorem falimd

Step	Hyp	Ref	Expression
1		falim 1328	. 2 ⊢ ( ⊥ → ψ)
2	1	adantl 452	1 ⊢ ((φ ∧ ⊥ ) → ψ)

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)