Theorem nanan 1493

Description: Conjunction in terms of alternative denial. (Contributed by Mario Carneiro, 9-May-2015.)

Assertion

Ref	Expression
nanan	⊢ ((𝜑 ∧ 𝜓) ↔ ¬ (𝜑 ⊼ 𝜓))

Proof of Theorem nanan

Step	Hyp	Ref	Expression
1		df-nan 1492	. 2 ⊢ ((𝜑 ⊼ 𝜓) ↔ ¬ (𝜑 ∧ 𝜓))
2	1	con2bii 357	1 ⊢ ((𝜑 ∧ 𝜓) ↔ ¬ (𝜑 ⊼ 𝜓))

Syntax hints:  ¬ wn 3   ↔ wb 206   ∧ wa 395   ⊼ wnan 1491

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 207  df-nan 1492

This theorem is referenced by:  nannan  1497  nanass  1510