Theorem mof 2588

Description: Version of df-mo 2564 with disjoint variable condition replaced by nonfreeness hypothesis. (Contributed by NM, 8-Mar-1995.) Extract dfmo 2565 from this proof, and prove mof 2588 from it (as of 30-Sep-2022, directly from df-mo 2564). (Revised by Wolf Lammen, 28-May-2019.) Avoid ax-13 2401. (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 2565	. 2 ⊢ (∃*𝑥𝜑 ↔ ∃𝑧∀𝑥(𝜑 → 𝑥 = 𝑧))
2		mof.1	. . . . 5 ⊢ Ⅎ𝑦𝜑
3		nfv 1946	. . . . 5 ⊢ Ⅎ𝑦 𝑥 = 𝑧
4	2, 3	nfim 1928	. . . 4 ⊢ Ⅎ𝑦(𝜑 → 𝑥 = 𝑧)
5	4	nfal 2353	. . 3 ⊢ Ⅎ𝑦∀𝑥(𝜑 → 𝑥 = 𝑧)
6		nfv 1946	. . 3 ⊢ Ⅎ𝑧∀𝑥(𝜑 → 𝑥 = 𝑦)
7		equequ2 2058	. . . . 5 ⊢ (𝑧 = 𝑦 → (𝑥 = 𝑧 ↔ 𝑥 = 𝑦))
8	7	imbi2d 343	. . . 4 ⊢ (𝑧 = 𝑦 → ((𝜑 → 𝑥 = 𝑧) ↔ (𝜑 → 𝑥 = 𝑦)))
9	8	albidv 1952	. . 3 ⊢ (𝑧 = 𝑦 → (∀𝑥(𝜑 → 𝑥 = 𝑧) ↔ ∀𝑥(𝜑 → 𝑥 = 𝑦)))
10	5, 6, 9	cbvexv1 2371	. 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 1567  ∃wex 1811  Ⅎwnf 1815  ∃*wmo 2562

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-10 2177  ax-11 2193  ax-12 2212

This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-ex 1812  df-nf 1816  df-mo 2564

This theorem is used by:  mo3  2589  mo  2590  rmo2  3833  dffun3f  9978  nmo  33133  fineqvrep  35838  bj-eu3f  37845  permaxrep  46089