Theorem abf 4367

Description: A class abstraction determined by a false formula is empty. (Contributed by NM, 20-Jan-2012.) Avoid ax-8 2147, ax-10 2178, ax-11 2194, ax-12 2215. (Revised by GG, 30-Jun-2024.)

Hypothesis

Ref	Expression
abf.1	⊢ ¬ 𝜑

Assertion

Ref	Expression
abf	⊢ {𝑥 ∣ 𝜑} = ∅

Proof of Theorem abf

Step	Hyp	Ref	Expression
1		abf.1	. . . 4 ⊢ ¬ 𝜑
2	1	bifal 1586	. . 3 ⊢ (𝜑 ↔ ⊥)
3	2	abbii 2829	. 2 ⊢ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥}
4		dfnul4 4284	. 2 ⊢ ∅ = {𝑥 ∣ ⊥}
5	3, 4	eqtr4i 2788	1 ⊢ {𝑥 ∣ 𝜑} = ∅

This proof depends on syntax axioms:  ¬ wn 3   = wceq 1570  ⊥wfal 1582  {cab 2740  ∅c0 4282

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-9 2155  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-fal 1583  df-ex 1813  df-sb 2100  df-clab 2741  df-cleq 2754  df-dif 3905  df-nul 4283

This theorem is used by:  mpo0  7502  0qs  8766  fi0  9394  join0  18497  meet0  18498  addsrid  28237  muls01  28385  mulsrid  28386  onaddscl  28550  onmulscl  28551  n0cut  28607  fmla0disjsuc  35985  pmapglb2xN  40653