Theorem ab0w 4342

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. Version of ab0 4343 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 4296	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
2	1	eqeq2i 2782	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
3		df-clab 2748	. . . . . 6 ⊢ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ [𝑦 / 𝑥]⊥)
4		sbv 2128	. . . . . 6 ⊢ ([𝑦 / 𝑥]⊥ ↔ ⊥)
5	3, 4	bitri 278	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ ⊥)
6	5	bibi2i 340	. . . 4 ⊢ ((𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ (𝜓 ↔ ⊥))
7	6	albii 1846	. . 3 ⊢ (∀𝑦(𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ ∀𝑦(𝜓 ↔ ⊥))
8		ab0w.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
9	8	eqabcbw 2843	. . 3 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑦(𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}))
10		nbfal 1582	. . . 4 ⊢ (¬ 𝜓 ↔ (𝜓 ↔ ⊥))
11	10	albii 1846	. . 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)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 209  ∀wal 1565   = wceq 1567  ⊥wfal 1579  [wsb 2097   ∈ wcel 2149  {cab 2747  ∅c0 4294

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1822  ax-4 1836  ax-5 1937  ax-6 1994  ax-7 2035  ax-9 2159  ax-ext 2741

This theorem depends on definitions:  df-bi 210  df-an 401  df-tru 1570  df-fal 1580  df-ex 1807  df-sb 2098  df-clab 2748  df-cleq 2761  df-dif 3916  df-nul 4295

This theorem is referenced by:  ab0orv  4346  rabeq0w  4351  relimasn  6088  0mpo0  7494  fsetdmprc0  8852