Theorem ab0w 4331

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. Version of ab0 4332 using implicit substitution, which requires fewer axioms. (Contributed by GG, 3-Oct-2024.)

Hypothesis

Ref	Expression
ab0w.1	⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
ab0w	⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑦 ¬ 𝜓)

Distinct variable groups:   𝑥,𝑦   𝜑,𝑦   𝜓,𝑥

Allowed substitution hints:   𝜑(𝑥)   𝜓(𝑦)

Proof of Theorem ab0w

Step	Hyp	Ref	Expression
1		dfnul4 4283	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
2	1	eqeq2i 2776	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
3		df-clab 2742	. . . . . 6 ⊢ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ [𝑦 / 𝑥]⊥)
4		sbv 2126	. . . . . 6 ⊢ ([𝑦 / 𝑥]⊥ ↔ ⊥)
5	3, 4	bitri 278	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ ⊥)
6	5	bibi2i 340	. . . 4 ⊢ ((𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ (𝜓 ↔ ⊥))
7	6	albii 1851	. . 3 ⊢ (∀𝑦(𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ ∀𝑦(𝜓 ↔ ⊥))
8		ab0w.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
9	8	eqabcbw 2838	. . 3 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑦(𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}))
10		nbfal 1584	. . . 4 ⊢ (¬ 𝜓 ↔ (𝜓 ↔ ⊥))
11	10	albii 1851	. . 3 ⊢ (∀𝑦 ¬ 𝜓 ↔ ∀𝑦(𝜓 ↔ ⊥))
12	7, 9, 11	3bitr4i 306	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑦 ¬ 𝜓)
13	2, 12	bitri 278	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑦 ¬ 𝜓)

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ↔ wb 209  ∀wal 1567   = wceq 1569  ⊥wfal 1581  [wsb 2100   ∈ wcel 2146  {cab 2741  ∅c0 4281

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1944  ax-6 2001  ax-7 2042  ax-9 2156  ax-ext 2735

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-fal 1582  df-ex 1812  df-sb 2101  df-clab 2742  df-cleq 2755  df-dif 4050  df-nul 4282

This theorem is used by:  ab0orv  4335  rabeq0w  4340  relimasn  6089  0mpo0  7513  fsetdmprc0  8891