Theorem nbfal 1555

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 1554	. 2 ⊢ ¬ ⊥
2	1	nbn 372	1 ⊢ (¬ 𝜑 ↔ (𝜑 ↔ ⊥))

Syntax hints:  ¬ wn 3   ↔ wb 206  ⊥wfal 1552

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 207  df-tru 1543  df-fal 1553

This theorem is referenced by:  nulmo  2712  eq0  4325  ab0  4355  eq0rdv  4382  rzal  4484  bisym1  36437  wl-1xor  37500  wl-1mintru1  37506  aisfina  46927  aifftbifffaibifff  46951  lindslinindsimp2  48439