Theorem mof 2589

Description: Version of df-mo 2565 with disjoint variable condition replaced by nonfreeness hypothesis. (Contributed by NM, 8-Mar-1995.) Extract dfmo 2566 from this proof, and prove mof 2589 from it (as of 30-Sep-2022, directly from df-mo 2565). (Revised by Wolf Lammen, 28-May-2019.) Avoid ax-13 2402. (Revised by Wolf Lammen, 16-Oct-2022.)

Hypothesis

Ref	Expression
mof.1	⊢ Ⅎ𝑦𝜑

Assertion

Ref	Expression
mof	⊢ (∃*𝑥𝜑 ↔ ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦))

Distinct variable group:   𝑥,𝑦

Allowed substitution hints:   𝜑(𝑥, 𝑦)

Proof of Theorem mof

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		dfmo 2566	. 2 ⊢ (∃*𝑥𝜑 ↔ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧))
2		mof.1	. . . . 5 ⊢ Ⅎ𝑦𝜑
3		nfv 1947	. . . . 5 ⊢ Ⅎ𝑦 𝑥 = 𝑧
4	2, 3	nfim 1929	. . . 4 ⊢ Ⅎ𝑦(𝜑 → 𝑥 = 𝑧)
5	4	nfal 2354	. . 3 ⊢ Ⅎ𝑦∀𝑥(𝜑 → 𝑥 = 𝑧)
6		nfv 1947	. . 3 ⊢ Ⅎ𝑧∀𝑥(𝜑 → 𝑥 = 𝑦)
7		equequ2 2059	. . . . 5 ⊢ (𝑧 = 𝑦 → (𝑥 = 𝑧 ↔ 𝑥 = 𝑦))
8	7	imbi2d 343	. . . 4 ⊢ (𝑧 = 𝑦 → ((𝜑 → 𝑥 = 𝑧) ↔ (𝜑 → 𝑥 = 𝑦)))
9	8	albidv 1953	. . 3 ⊢ (𝑧 = 𝑦 → (∀𝑥(𝜑 → 𝑥 = 𝑧) ↔ ∀𝑥(𝜑 → 𝑥 = 𝑦)))
10	5, 6, 9	cbvexv1 2372	. 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:   → wi 4   ↔ wb 209  ∀wal 1568  ∃wex 1812  Ⅎwnf 1816  ∃*wmo 2563

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-10 2178  ax-11 2194  ax-12 2213

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1813  df-nf 1817  df-mo 2565

This theorem is used by:  mo3  2590  mo  2591  rmo2  3834  dffun3f  9956  nmo  33068  fineqvrep  35755  bj-eu3f  37723  permaxrep  45948