Metamath Proof Explorer


Theorem efald

Description: Deduction based on reductio ad absurdum. (Contributed by Mario Carneiro, 9-Feb-2017)

Ref Expression
Hypothesis efald.1 ( ( 𝜑 ∧ ¬ 𝜓 ) → ⊥ )
Assertion efald ( 𝜑𝜓 )

Proof

Step Hyp Ref Expression
1 efald.1 ( ( 𝜑 ∧ ¬ 𝜓 ) → ⊥ )
2 1 inegd ( 𝜑 → ¬ ¬ 𝜓 )
3 2 notnotrd ( 𝜑𝜓 )