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Theorem eqreu 3690
Description: A condition which implies existential uniqueness. (Contributed by Mario Carneiro, 2-Oct-2015.)
Hypothesis
Ref Expression
eqreu.1 (𝑥 = 𝐵 → (𝜑 ↔ 𝜓))
Assertion
Ref Expression
eqreu ((𝐵 ∈ 𝐴 ∧ 𝜓 ∧ ∀𝑥 ∈ 𝐴 (𝜑 → 𝑥 = 𝐵)) → ∃!𝑥 ∈ 𝐴 𝜑)
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵   𝜓,𝑥
Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem eqreu
StepHypRef Expression
1 ralbiim 3128 . . . . 5 (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝑥 = 𝐵) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝑥 = 𝐵) ∧ ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 → 𝜑)))
2 eqreu.1 . . . . . . 7 (𝑥 = 𝐵 → (𝜑 ↔ 𝜓))
32ceqsralv 3494 . . . . . 6 (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 → 𝜑) ↔ 𝜓))
43anbi2d 642 . . . . 5 (𝐵 ∈ 𝐴 → ((∀𝑥 ∈ 𝐴 (𝜑 → 𝑥 = 𝐵) ∧ ∀𝑥 ∈ 𝐴 (𝑥 = 𝐵 → 𝜑)) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝑥 = 𝐵) ∧ 𝜓)))
51, 4bitrid 286 . . . 4 (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝑥 = 𝐵) ↔ (∀𝑥 ∈ 𝐴 (𝜑 → 𝑥 = 𝐵) ∧ 𝜓)))
6 reu6i 3689 . . . . 5 ((𝐵 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝑥 = 𝐵)) → ∃!𝑥 ∈ 𝐴 𝜑)
76ex 418 . . . 4 (𝐵 ∈ 𝐴 → (∀𝑥 ∈ 𝐴 (𝜑 ↔ 𝑥 = 𝐵) → ∃!𝑥 ∈ 𝐴 𝜑))
85, 7sylbird 263 . . 3 (𝐵 ∈ 𝐴 → ((∀𝑥 ∈ 𝐴 (𝜑 → 𝑥 = 𝐵) ∧ 𝜓) → ∃!𝑥 ∈ 𝐴 𝜑))
983impib 1134 . 2 ((𝐵 ∈ 𝐴 ∧ ∀𝑥 ∈ 𝐴 (𝜑 → 𝑥 = 𝐵) ∧ 𝜓) → ∃!𝑥 ∈ 𝐴 𝜑)
1093com23 1144 1 ((𝐵 ∈ 𝐴 ∧ 𝜓 ∧ ∀𝑥 ∈ 𝐴 (𝜑 → 𝑥 = 𝐵)) → ∃!𝑥 ∈ 𝐴 𝜑)
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401   ∧ w3a 1103   = wceq 1570   ∈ wcel 2147  ∀wral 3080  ∃!wreu 3367
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 1945  ax-6 2002  ax-7 2043  ax-8 2149  ax-9 2157  ax-10 2180  ax-12 2216  ax-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-or 862  df-3an 1105  df-tru 1573  df-ex 1813  df-nf 1817  df-sb 2102  df-mo 2567  df-eu 2597  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-rex 3091  df-reu 3370
This theorem is used by:  uzwo3  13110  frmdup3  19113  frgpup3  20043  neiptopreu  23506  ufileu  24293  mirreu  29193  lmireu  29352  opreu2reuALT  33172  ccatws1f1o  33610  symgfcoeu  33742  aks6d1c7lem4  43499
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