Theorem ab0w 4379

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. Version of ab0 4380 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 4335	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
2	1	eqeq2i 2750	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
3		dfcleq 2730	. . . 4 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ ⊥}))
4		df-clab 2715	. . . . . . . 8 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ [𝑦 / 𝑥]𝜑)
5		ab0w.1	. . . . . . . . 9 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
6	5	sbievw 2093	. . . . . . . 8 ⊢ ([𝑦 / 𝑥]𝜑 ↔ 𝜓)
7	4, 6	bitri 275	. . . . . . 7 ⊢ (𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝜓)
8	7	bibi1i 338	. . . . . 6 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ (𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}))
9		bicom 222	. . . . . 6 ⊢ ((𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ 𝜓))
10	8, 9	bitri 275	. . . . 5 ⊢ ((𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ 𝜓))
11	10	albii 1819	. . . 4 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ 𝜑} ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊥} ↔ 𝜓))
12	3, 11	bitri 275	. . 3 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑦(𝑦 ∈ {𝑥 ∣ ⊥} ↔ 𝜓))
13		df-clab 2715	. . . . . . 7 ⊢ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ [𝑦 / 𝑥]⊥)
14		sbv 2088	. . . . . . 7 ⊢ ([𝑦 / 𝑥]⊥ ↔ ⊥)
15	13, 14	bitri 275	. . . . . 6 ⊢ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ ⊥)
16	15	bibi1i 338	. . . . 5 ⊢ ((𝑦 ∈ {𝑥 ∣ ⊥} ↔ 𝜓) ↔ (⊥ ↔ 𝜓))
17	16	albii 1819	. . . 4 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ ⊥} ↔ 𝜓) ↔ ∀𝑦(⊥ ↔ 𝜓))
18		falim 1557	. . . . . . 7 ⊢ (⊥ → (𝜓 → ¬ 𝜓))
19		idd 24	. . . . . . 7 ⊢ (¬ ⊥ → (¬ 𝜓 → ¬ 𝜓))
20	18, 19	bija 380	. . . . . 6 ⊢ ((⊥ ↔ 𝜓) → ¬ 𝜓)
21		falim 1557	. . . . . . 7 ⊢ (⊥ → 𝜓)
22		id 22	. . . . . . 7 ⊢ (𝜓 → 𝜓)
23	21, 22	pm5.21ni 377	. . . . . 6 ⊢ (¬ 𝜓 → (⊥ ↔ 𝜓))
24	20, 23	impbii 209	. . . . 5 ⊢ ((⊥ ↔ 𝜓) ↔ ¬ 𝜓)
25	24	albii 1819	. . . 4 ⊢ (∀𝑦(⊥ ↔ 𝜓) ↔ ∀𝑦 ¬ 𝜓)
26	17, 25	bitri 275	. . 3 ⊢ (∀𝑦(𝑦 ∈ {𝑥 ∣ ⊥} ↔ 𝜓) ↔ ∀𝑦 ¬ 𝜓)
27	12, 26	bitri 275	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑦 ¬ 𝜓)
28	2, 27	bitri 275	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑦 ¬ 𝜓)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 206  ∀wal 1538   = wceq 1540  ⊥wfal 1552  [wsb 2064   ∈ wcel 2108  {cab 2714  ∅c0 4333

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2007  ax-9 2118  ax-ext 2708

This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1543  df-fal 1553  df-ex 1780  df-sb 2065  df-clab 2715  df-cleq 2729  df-dif 3954  df-nul 4334

This theorem is referenced by:  rabeq0w  4387  0mpo0  7516  fsetdmprc0  8895