Theorem cbvreuvw 2792

Description: Version of cbvreuv 2788 with a disjoint variable condition. (Contributed by GG, 10-Jan-2024.) Reduce axiom usage. (Revised by GG, 25-Aug-2024.)

Hypothesis

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

Assertion

Ref	Expression
cbvreuvw	⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑦 ∈ 𝐴 𝜓)

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

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

Proof of Theorem cbvreuvw

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eleq1w 2299	. . . . . . 7 ⊢ (𝑥 = 𝑦 → (𝑥 ∈ 𝐴 ↔ 𝑦 ∈ 𝐴))
2		cbvralvw.1	. . . . . . 7 ⊢ (𝑥 = 𝑦 → (𝜑 ↔ 𝜓))
3	1, 2	anbi12d 477	. . . . . 6 ⊢ (𝑥 = 𝑦 → ((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ (𝑦 ∈ 𝐴 ∧ 𝜓)))
4		equequ1 1764	. . . . . 6 ⊢ (𝑥 = 𝑦 → (𝑥 = 𝑧 ↔ 𝑦 = 𝑧))
5	3, 4	bibi12d 235	. . . . 5 ⊢ (𝑥 = 𝑦 → (((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ 𝑥 = 𝑧) ↔ ((𝑦 ∈ 𝐴 ∧ 𝜓) ↔ 𝑦 = 𝑧)))
6	5	cbvalvw 1975	. . . 4 ⊢ (∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ 𝑥 = 𝑧) ↔ ∀𝑦((𝑦 ∈ 𝐴 ∧ 𝜓) ↔ 𝑦 = 𝑧))
7	6	exbii 1658	. . 3 ⊢ (∃𝑧∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ 𝑥 = 𝑧) ↔ ∃𝑧∀𝑦((𝑦 ∈ 𝐴 ∧ 𝜓) ↔ 𝑦 = 𝑧))
8		df-eu 2089	. . 3 ⊢ (∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑) ↔ ∃𝑧∀𝑥((𝑥 ∈ 𝐴 ∧ 𝜑) ↔ 𝑥 = 𝑧))
9		df-eu 2089	. . 3 ⊢ (∃!𝑦(𝑦 ∈ 𝐴 ∧ 𝜓) ↔ ∃𝑧∀𝑦((𝑦 ∈ 𝐴 ∧ 𝜓) ↔ 𝑦 = 𝑧))
10	7, 8, 9	3bitr4ri 213	. 2 ⊢ (∃!𝑦(𝑦 ∈ 𝐴 ∧ 𝜓) ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
11		df-reu 2535	. 2 ⊢ (∃!𝑦 ∈ 𝐴 𝜓 ↔ ∃!𝑦(𝑦 ∈ 𝐴 ∧ 𝜓))
12		df-reu 2535	. 2 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑥(𝑥 ∈ 𝐴 ∧ 𝜑))
13	10, 11, 12	3bitr4ri 213	1 ⊢ (∃!𝑥 ∈ 𝐴 𝜑 ↔ ∃!𝑦 ∈ 𝐴 𝜓)

This proof depends on syntax axioms:   → wi 4   ∧ wa 104   ↔ wb 105  ∀wal 1400  ∃wex 1545  ∃!weu 2086   ∈ wcel 2209  ∃!wreu 2530

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1500  ax-gen 1502  ax-ie1 1546  ax-ie2 1547  ax-8 1557  ax-4 1563  ax-17 1579  ax-i9 1583  ax-ial 1587

This proof depends on definitions:  df-bi 117  df-nf 1514  df-eu 2089  df-clel 2234  df-reu 2535

This theorem is used by: (None)