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Theorem sbeqalb 3800
Description: Theorem *14.121 in [WhiteheadRussell] p. 185. (Contributed by Andrew Salmon, 28-Jun-2011.) (Proof shortened by Wolf Lammen, 9-May-2013.)
Assertion
Ref Expression
sbeqalb (𝐴 ∈ 𝑉 → ((∀𝑥(𝜑 ↔ 𝑥 = 𝐴) ∧ ∀𝑥(𝜑 ↔ 𝑥 = 𝐵)) → 𝐴 = 𝐵))
Distinct variable groups:   𝑥,𝐴   𝑥,𝐵
Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem sbeqalb
StepHypRef Expression
1 bibi1 354 . . . . 5 ((𝜑 ↔ 𝑥 = 𝐴) → ((𝜑 ↔ 𝑥 = 𝐵) ↔ (𝑥 = 𝐴 ↔ 𝑥 = 𝐵)))
21biimpa 482 . . . 4 (((𝜑 ↔ 𝑥 = 𝐴) ∧ (𝜑 ↔ 𝑥 = 𝐵)) → (𝑥 = 𝐴 ↔ 𝑥 = 𝐵))
32biimpd 232 . . 3 (((𝜑 ↔ 𝑥 = 𝐴) ∧ (𝜑 ↔ 𝑥 = 𝐵)) → (𝑥 = 𝐴 → 𝑥 = 𝐵))
43alanimi 1848 . 2 ((∀𝑥(𝜑 ↔ 𝑥 = 𝐴) ∧ ∀𝑥(𝜑 ↔ 𝑥 = 𝐵)) → ∀𝑥(𝑥 = 𝐴 → 𝑥 = 𝐵))
5 sbceqal 3799 . 2 (𝐴 ∈ 𝑉 → (∀𝑥(𝑥 = 𝐴 → 𝑥 = 𝐵) → 𝐴 = 𝐵))
64, 5syl5 35 1 (𝐴 ∈ 𝑉 → ((∀𝑥(𝜑 ↔ 𝑥 = 𝐴) ∧ ∀𝑥(𝜑 ↔ 𝑥 = 𝐵)) → 𝐴 = 𝐵))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209   ∧ wa 401  ∀wal 1567   = wceq 1569   ∈ wcel 2144
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-8 2146  ax-9 2154  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-v 3452
This theorem is used by: (None)
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