Theorem eqeu 3699

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 3599	. . . 4 ⊢ (𝐴 ∈ 𝐵 → (𝜓 → ∃𝑥𝜑))
3	2	imp 409	. . 3 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝜓) → ∃𝑥𝜑)
4	3	3adant3 1128	. 2 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝜓 ∧ ∀𝑥(𝜑 → 𝑥 = 𝐴)) → ∃𝑥𝜑)
5		eqeq2 2835	. . . . . . 7 ⊢ (𝑦 = 𝐴 → (𝑥 = 𝑦 ↔ 𝑥 = 𝐴))
6	5	imbi2d 343	. . . . . 6 ⊢ (𝑦 = 𝐴 → ((𝜑 → 𝑥 = 𝑦) ↔ (𝜑 → 𝑥 = 𝐴)))
7	6	albidv 1921	. . . . 5 ⊢ (𝑦 = 𝐴 → (∀𝑥(𝜑 → 𝑥 = 𝑦) ↔ ∀𝑥(𝜑 → 𝑥 = 𝐴)))
8	7	spcegv 3599	. . . 4 ⊢ (𝐴 ∈ 𝐵 → (∀𝑥(𝜑 → 𝑥 = 𝐴) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))
9	8	imp 409	. . 3 ⊢ ((𝐴 ∈ 𝐵 ∧ ∀𝑥(𝜑 → 𝑥 = 𝐴)) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦))
10	9	3adant2 1127	. 2 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝜓 ∧ ∀𝑥(𝜑 → 𝑥 = 𝐴)) → ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦))
11		eu3v 2655	. 2 ⊢ (∃!𝑥𝜑 ↔ (∃𝑥𝜑 ∧ ∃𝑦∀𝑥(𝜑 → 𝑥 = 𝑦)))
12	4, 10, 11	sylanbrc 585	1 ⊢ ((𝐴 ∈ 𝐵 ∧ 𝜓 ∧ ∀𝑥(𝜑 → 𝑥 = 𝐴)) → ∃!𝑥𝜑)

Syntax hints:   → wi 4   ↔ wb 208   ∧ w3a 1083  ∀wal 1535   = wceq 1537  ∃wex 1780   ∈ wcel 2114  ∃!weu 2653

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1796  ax-4 1810  ax-5 1911  ax-6 1970  ax-7 2015  ax-8 2116  ax-9 2124  ax-ext 2795

This theorem depends on definitions:  df-bi 209  df-an 399  df-3an 1085  df-ex 1781  df-mo 2622  df-eu 2654  df-cleq 2816  df-clel 2895

This theorem is referenced by:  rngurd  30859  neibastop3  33712  upixp  35006  zrinitorngc  44278  zrtermorngc  44279  zrtermoringc  44348