Theorem nbfal 1310

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 1306	. 2 ⊢ ¬ ⊥
2	1	nbn 656	1 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))

Syntax hints:  ¬ wn 3   ↔ wb 104  ⊥wfal 1304

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-in1 584  ax-in2 585

This theorem depends on definitions:  df-bi 116  df-tru 1302  df-fal 1305

This theorem is referenced by:  zfnuleu  3992