Theorem nbfal 1556

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 1555	. 2 ⊢ ¬ ⊥
2	1	nbn 372	1 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))

Syntax hints:  ¬ wn 3   ↔ wb 205  ⊥wfal 1553

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 206  df-tru 1544  df-fal 1554

This theorem is referenced by:  nulmo  2715  eq0  4282  ab0  4313  ab0OLD  4314  eq0rdv  4343  rzal  4444  bisym1  34587  wl-1xor  35632  wl-1mintru1  35638  aisfina  44344  aifftbifffaibifff  44368  lindslinindsimp2  45756