Theorem nbfal 1406

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 1402	. 2 ⊢ ¬ ⊥
2	1	nbn 704	1 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))

Syntax hints:  ¬ wn 3   ↔ wb 105  ⊥wfal 1400

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 617  ax-in2 618

This theorem depends on definitions:  df-bi 117  df-tru 1398  df-fal 1401

This theorem is referenced by:  zfnuleu  4207