Theorem euf 2603

Description: Version of eu6 2601 with disjoint variable condition replaced by nonfreeness hypothesis. (Contributed by NM, 12-Aug-1993.) (Proof shortened by Wolf Lammen, 30-Oct-2018.) Avoid ax-13 2403. (Revised by Wolf Lammen, 16-Oct-2022.)

Hypothesis

Ref	Expression
euf.1	⊢ Ⅎ𝑦𝜑

Assertion

Ref	Expression
euf	⊢ (∃!𝑥𝜑 ↔ ∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)

Proof of Theorem euf

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eu6 2601	. 2 ⊢ (∃!𝑥𝜑 ↔ ∃𝑧∀𝑥(𝜑 ↔ 𝑥 = 𝑧))
2		euf.1	. . . . 5 ⊢ Ⅎ𝑦𝜑
3		nfv 1943	. . . . 5 ⊢ Ⅎ𝑦 𝑥 = 𝑧
4	2, 3	nfbi 1932	. . . 4 ⊢ Ⅎ𝑦(𝜑 ↔ 𝑥 = 𝑧)
5	4	nfal 2355	. . 3 ⊢ Ⅎ𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑧)
6		nfv 1943	. . 3 ⊢ Ⅎ𝑧∀𝑥(𝜑 ↔ 𝑥 = 𝑦)
7		equequ2 2055	. . . . 5 ⊢ (𝑧 = 𝑦 → (𝑥 = 𝑧 ↔ 𝑥 = 𝑦))
8	7	bibi2d 345	. . . 4 ⊢ (𝑧 = 𝑦 → ((𝜑 ↔ 𝑥 = 𝑧) ↔ (𝜑 ↔ 𝑥 = 𝑦)))
9	8	albidv 1949	. . 3 ⊢ (𝑧 = 𝑦 → (∀𝑥(𝜑 ↔ 𝑥 = 𝑧) ↔ ∀𝑥(𝜑 ↔ 𝑥 = 𝑦)))
10	5, 6, 9	cbvexv1 2373	. 2 ⊢ (∃𝑧∀𝑥(𝜑 ↔ 𝑥 = 𝑧) ↔ ∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦))
11	1, 10	bitri 278	1 ⊢ (∃!𝑥𝜑 ↔ ∃𝑦∀𝑥(𝜑 ↔ 𝑥 = 𝑦))

This proof depends on syntax axioms:   ↔ wb 209  ∀wal 1567  ∃wex 1808  Ⅎwnf 1812  ∃!weu 2595

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-10 2175  ax-11 2191  ax-12 2212

This proof depends on definitions:  df-bi 210  df-an 401  df-or 861  df-tru 1572  df-ex 1809  df-nf 1813  df-mo 2566  df-eu 2596

This theorem is used by:  eu1  2637