Theorem nbfal 1552

Description: The negation of a proposition is equivalent to itself being equivalent to ⊥. (Contributed by Anthony Hart, 14-Aug-2011.)

Assertion

Ref	Expression
nbfal	⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))

Proof of Theorem nbfal

Step	Hyp	Ref	Expression
1		fal 1551	. 2 ⊢ ¬ ⊥
2	1	nbn 375	1 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))

Syntax hints:  ¬ wn 3   ↔ wb 208  ⊥wfal 1549

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-tru 1540  df-fal 1550

This theorem is referenced by:  nulmo  2798  bisym1  33767  aisfina  43154  aifftbifffaibifff  43178  lindslinindsimp2  44538