Theorem bj-abf 37572

Description: Shorter proof of abf 4370 (which should be kept as abfALT). (Contributed by BJ, 24-Jul-2019.) (Proof modification is discouraged.)

Hypothesis

Ref	Expression
bj-abf.1	⊢ ¬ 𝜑

Assertion

Ref	Expression
bj-abf	⊢ {𝑥 ∣ 𝜑} = ∅

Proof of Theorem bj-abf

Step	Hyp	Ref	Expression
1		bj-ab0 37571	. 2 ⊢ (∀𝑥 ¬ 𝜑 → {𝑥 ∣ 𝜑} = ∅)
2		bj-abf.1	. 2 ⊢ ¬ 𝜑
3	1, 2	mpg 1826	1 ⊢ {𝑥 ∣ 𝜑} = ∅

This proof depends on syntax axioms:  ¬ wn 3   = wceq 1569  {cab 2740  ∅c0 4285

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-9 2152  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-tru 1572  df-fal 1582  df-ex 1809  df-sb 2096  df-clab 2741  df-cleq 2754  df-dif 3907  df-nul 4286

This theorem is used by: (None)