Theorem eqeu 2900

Description: A condition which implies existential uniqueness. (Contributed by Jeff Hankins, 8-Sep-2009.)

Hypothesis

Ref	Expression
eqeu.1	⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))

Assertion

Ref	Expression
eqeu	⊢ ((𝐴 ∈ 𝐵 ∧ 𝜓 ∧ ∀𝑥(𝜑 → 𝑥 = 𝐴)) → ∃!𝑥𝜑)

Distinct variable groups:   𝜓,𝑥   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥)   𝐵(𝑥)

Proof of Theorem eqeu

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		eqeu.1	. . . . 5 ⊢ (𝑥 = 𝐴 → (𝜑 ↔ 𝜓))
2	1	spcegv 2818	. . . 4 ⊢ (𝐴 ∈ 𝐵 → (𝜓 → ∃𝑥𝜑))
3	2	imp 123	. . 3 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝜓) → ∃𝑥𝜑)
4	3	3adant3 1012	. 2 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝜓 ∧ ∀𝑥(𝜑 → 𝑥 = 𝐴)) → ∃𝑥𝜑)
5		eqeq2 2180	. . . . . . 7 ⊢ (𝑦 = 𝐴 → (𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
6	5	imbi2d 229	. . . . . 6 ⊢ (𝑦 = 𝐴 → ((𝜑 → 𝑥 = 𝑦) ↔ (𝜑 → 𝑥 = 𝐴)))
7	6	albidv 1817	. . . . 5 ⊢ (𝑦 = 𝐴 → (∀𝑥(𝜑 → 𝑥 = 𝑦) ↔ ∀𝑥(𝜑 → 𝑥 = 𝐴)))
8	7	spcegv 2818	. . . 4 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥(𝜑 → 𝑥 = 𝐴) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))
9	8	imp 123	. . 3 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥(𝜑 → 𝑥 = 𝐴)) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦))
10	9	3adant2 1011	. 2 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝜓 ∧ ∀𝑥(𝜑 → 𝑥 = 𝐴)) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦))
11		nfv 1521	. . 3 ⊢ Ⅎ𝑦𝜑
12	11	eu3 2065	. 2 ⊢ (∃!𝑥𝜑 ↔ (∃𝑥𝜑 ∧ ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))
13	4, 10, 12	sylanbrc 415	1 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝜓 ∧ ∀𝑥(𝜑 → 𝑥 = 𝐴)) → ∃!𝑥𝜑)

Syntax hints:   → wi 4   ↔ wb 104   ∧ w3a 973  ∀wal 1346   = wceq 1348  ∃wex 1485  ∃!weu 2019   ∈ wcel 2141

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 704  ax-5 1440  ax-7 1441  ax-gen 1442  ax-ie1 1486  ax-ie2 1487  ax-8 1497  ax-10 1498  ax-11 1499  ax-i12 1500  ax-bndl 1502  ax-4 1503  ax-17 1519  ax-i9 1523  ax-ial 1527  ax-i5r 1528  ax-ext 2152

This theorem depends on definitions:  df-bi 116  df-3an 975  df-tru 1351  df-nf 1454  df-sb 1756  df-eu 2022  df-clab 2157  df-cleq 2163  df-clel 2166  df-nfc 2301  df-v 2732

This theorem is referenced by: (None)