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Theorem pm13.183 3619
Description: Compare theorem *13.183 in [WhiteheadRussell] p. 178. Only 𝐴 is required to be a set. (Contributed by Andrew Salmon, 3-Jun-2011.) Avoid ax-13 2401. (Revised by Wolf Lammen, 29-Apr-2023.)
Assertion
Ref Expression
pm13.183 (𝐴 ∈ 𝑉 → (𝐴 = 𝐵 ↔ ∀𝑧(𝑧 = 𝐴 ↔ 𝑧 = 𝐵)))
Distinct variable groups:   𝑧,𝐴   𝑧,𝐵
Allowed substitution hint:   𝑉(𝑧)

Proof of Theorem pm13.183
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 eqeq1 2764 . 2 (𝑦 = 𝐴 → (𝑦 = 𝐵 ↔ 𝐴 = 𝐵))
2 eqeq2 2772 . . . 4 (𝑦 = 𝐴 → (𝑧 = 𝑦 ↔ 𝑧 = 𝐴))
32bibi1d 346 . . 3 (𝑦 = 𝐴 → ((𝑧 = 𝑦 ↔ 𝑧 = 𝐵) ↔ (𝑧 = 𝐴 ↔ 𝑧 = 𝐵)))
43albidv 1952 . 2 (𝑦 = 𝐴 → (∀𝑧(𝑧 = 𝑦 ↔ 𝑧 = 𝐵) ↔ ∀𝑧(𝑧 = 𝐴 ↔ 𝑧 = 𝐵)))
5 eqeq2 2772 . . . 4 (𝑦 = 𝐵 → (𝑧 = 𝑦 ↔ 𝑧 = 𝐵))
65alrimiv 1959 . . 3 (𝑦 = 𝐵 → ∀𝑧(𝑧 = 𝑦 ↔ 𝑧 = 𝐵))
7 stdpc4 2104 . . . 4 (∀𝑧(𝑧 = 𝑦 ↔ 𝑧 = 𝐵) → [𝑦 / 𝑧](𝑧 = 𝑦 ↔ 𝑧 = 𝐵))
8 sbbi 2340 . . . . 5 ([𝑦 / 𝑧](𝑧 = 𝑦 ↔ 𝑧 = 𝐵) ↔ ([𝑦 / 𝑧]𝑧 = 𝑦 ↔ [𝑦 / 𝑧]𝑧 = 𝐵))
9 equsb1v 2141 . . . . . . 7 [𝑦 / 𝑧]𝑧 = 𝑦
109tbt 372 . . . . . 6 ([𝑦 / 𝑧]𝑧 = 𝐵 ↔ ([𝑦 / 𝑧]𝑧 = 𝐵 ↔ [𝑦 / 𝑧]𝑧 = 𝑦))
11 bicom 225 . . . . . 6 (([𝑦 / 𝑧]𝑧 = 𝐵 ↔ [𝑦 / 𝑧]𝑧 = 𝑦) ↔ ([𝑦 / 𝑧]𝑧 = 𝑦 ↔ [𝑦 / 𝑧]𝑧 = 𝐵))
1210, 11bitri 278 . . . . 5 ([𝑦 / 𝑧]𝑧 = 𝐵 ↔ ([𝑦 / 𝑧]𝑧 = 𝑦 ↔ [𝑦 / 𝑧]𝑧 = 𝐵))
13 eqsb1 2886 . . . . 5 ([𝑦 / 𝑧]𝑧 = 𝐵 ↔ 𝑦 = 𝐵)
148, 12, 133bitr2i 302 . . . 4 ([𝑦 / 𝑧](𝑧 = 𝑦 ↔ 𝑧 = 𝐵) ↔ 𝑦 = 𝐵)
157, 14sylib 221 . . 3 (∀𝑧(𝑧 = 𝑦 ↔ 𝑧 = 𝐵) → 𝑦 = 𝐵)
166, 15impbii 212 . 2 (𝑦 = 𝐵 ↔ ∀𝑧(𝑧 = 𝑦 ↔ 𝑧 = 𝐵))
171, 4, 16vtoclbg 3519 1 (𝐴 ∈ 𝑉 → (𝐴 = 𝐵 ↔ ∀𝑧(𝑧 = 𝐴 ↔ 𝑧 = 𝐵)))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567   = wceq 1569  [wsb 2098   ∈ 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-10 2177  ax-12 2212  ax-ext 2732
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-nf 1816  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835
This theorem is used by:  mpo2eqb  7552
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