Theorem nbfal 1585

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 1584	. 2 ⊢ ¬ ⊥
2	1	nbn 375	1 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))

This proof depends on syntax axioms:  ¬ wn 3   ↔ wb 209  ⊥wfal 1582

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-tru 1573  df-fal 1583

This theorem is used by:  nulmo  2742  eq0  4304  ab0w  4335  ab0  4336  bisym1  36989  wl-1xor  38187  wl-1mintru1  38193  aisfina  47695  aifftbifffaibifff  47719  lindslinindsimp2  49302