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Theorem nanan 1483
Description: Conjunction in terms of alternative denial. (Contributed by Mario Carneiro, 9-May-2015.)
Assertion
Ref Expression
nanan ((𝜑𝜓) ↔ ¬ (𝜑𝜓))

Proof of Theorem nanan
StepHypRef Expression
1 df-nan 1482 . 2 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
21con2bii 360 1 ((𝜑𝜓) ↔ ¬ (𝜑𝜓))
Colors of variables: wff setvar class
Syntax hints:  ¬ wn 3  wb 208  wa 398  wnan 1481
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 209  df-nan 1482
This theorem is referenced by:  nannan  1487  nanass  1500
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