Theorem ab0w 3550

Description: The class of sets verifying a property is the empty class if and only if that property is a contradiction. (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 3522	. . 3 ⊢ ∅ = {𝑥 ∣ ⊥}
2	1	eqeq2i 2249	. 2 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ {𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥})
3		df-clab 2225	. . . . . 6 ⊢ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ [𝑦 / 𝑥]⊥)
4		sbv 1949	. . . . . 6 ⊢ ([𝑦 / 𝑥]⊥ ↔ ⊥)
5	3, 4	bitri 184	. . . . 5 ⊢ (𝑦 ∈ {𝑥 ∣ ⊥} ↔ ⊥)
6	5	bibi2i 227	. . . 4 ⊢ ((𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ (𝜓 ↔ ⊥))
7	6	albii 1523	. . 3 ⊢ (∀𝑦(𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}) ↔ ∀𝑦(𝜓 ↔ ⊥))
8		ab0w.1	. . . 4 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
9	8	eqabcbw 2376	. . 3 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑦(𝜓 ↔ 𝑦 ∈ {𝑥 ∣ ⊥}))
10		nbfal 1413	. . . 4 ⊢ (¬ 𝜓 ↔ (𝜓 ↔ ⊥))
11	10	albii 1523	. . 3 ⊢ (∀𝑦 ¬ 𝜓 ↔ ∀𝑦(𝜓 ↔ ⊥))
12	7, 9, 11	3bitr4i 212	. 2 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ ⊥} ↔ ∀𝑦 ¬ 𝜓)
13	2, 12	bitri 184	1 ⊢ ({𝑥 ∣ 𝜑} = ∅ ↔ ∀𝑦 ¬ 𝜓)

Syntax hints:  ¬ wn 3   → wi 4   ↔ wb 105  ∀wal 1400   = wceq 1402  ⊥wfal 1407  [wsb 1815   ∈ wcel 2209  {cab 2224  ∅c0 3520

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-in1 623  ax-in2 624  ax-io 721  ax-5 1500  ax-7 1501  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-10 1558  ax-11 1559  ax-i12 1560  ax-bndl 1562  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587  ax-i5r 1588  ax-ext 2220

This theorem depends on definitions:  df-bi 117  df-tru 1405  df-fal 1408  df-nf 1514  df-sb 1816  df-clab 2225  df-cleq 2231  df-clel 2234  df-dif 3222  df-nul 3521

This theorem is referenced by:  fsetdmprc0  6940