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Theorem ceqex 3655
Description: Equality implies equivalence with substitution. (Contributed by NM, 2-Mar-1995.) (Proof shortened by BJ, 1-May-2019.)
Assertion
Ref Expression
ceqex (𝑥 = 𝐴 → (𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
Distinct variable group:   𝑥,𝐴
Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem ceqex
StepHypRef Expression
1 19.8a 2181 . . 3 ((𝑥 = 𝐴𝜑) → ∃𝑥(𝑥 = 𝐴𝜑))
21ex 412 . 2 (𝑥 = 𝐴 → (𝜑 → ∃𝑥(𝑥 = 𝐴𝜑)))
3 eqvisset 3501 . . . 4 (𝑥 = 𝐴𝐴 ∈ V)
4 alexeqg 3654 . . . 4 (𝐴 ∈ V → (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
53, 4syl 17 . . 3 (𝑥 = 𝐴 → (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
6 sp 2183 . . . 4 (∀𝑥(𝑥 = 𝐴𝜑) → (𝑥 = 𝐴𝜑))
76com12 32 . . 3 (𝑥 = 𝐴 → (∀𝑥(𝑥 = 𝐴𝜑) → 𝜑))
85, 7sylbird 260 . 2 (𝑥 = 𝐴 → (∃𝑥(𝑥 = 𝐴𝜑) → 𝜑))
92, 8impbid 212 1 (𝑥 = 𝐴 → (𝜑 ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
Colors of variables: wff setvar class
Syntax hints:  wi 4  wb 206  wa 395  wal 1537   = wceq 1539  wex 1778  wcel 2108  Vcvv 3481
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1794  ax-4 1808  ax-5 1910  ax-6 1967  ax-7 2007  ax-8 2110  ax-9 2118  ax-10 2141  ax-12 2177  ax-ext 2708
This theorem depends on definitions:  df-bi 207  df-an 396  df-tru 1542  df-ex 1779  df-nf 1783  df-sb 2065  df-clab 2715  df-cleq 2729  df-clel 2816  df-v 3483
This theorem is referenced by:  ceqsexg  3656
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