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Theorem alexeq 2901
Description: Two ways to express substitution of 𝐴 for 𝑥 in 𝜑. (Contributed by NM, 2-Mar-1995.)
Hypothesis
Ref Expression
alexeq.1 𝐴 ∈ V
Assertion
Ref Expression
alexeq (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑))
Distinct variable group:   𝑥,𝐴
Allowed substitution hint:   𝜑(𝑥)

Proof of Theorem alexeq
Dummy variable 𝑦 is distinct from all other variables.
StepHypRef Expression
1 alexeq.1 . . 3 𝐴 ∈ V
2 eqeq2 2216 . . . . 5 (𝑦 = 𝐴 → (𝑥 = 𝑦𝑥 = 𝐴))
32anbi1d 465 . . . 4 (𝑦 = 𝐴 → ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝐴𝜑)))
43exbidv 1849 . . 3 (𝑦 = 𝐴 → (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑)))
52imbi1d 231 . . . 4 (𝑦 = 𝐴 → ((𝑥 = 𝑦𝜑) ↔ (𝑥 = 𝐴𝜑)))
65albidv 1848 . . 3 (𝑦 = 𝐴 → (∀𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝐴𝜑)))
7 sb56 1910 . . 3 (∃𝑥(𝑥 = 𝑦𝜑) ↔ ∀𝑥(𝑥 = 𝑦𝜑))
81, 4, 6, 7vtoclb 2832 . 2 (∃𝑥(𝑥 = 𝐴𝜑) ↔ ∀𝑥(𝑥 = 𝐴𝜑))
98bicomi 132 1 (∀𝑥(𝑥 = 𝐴𝜑) ↔ ∃𝑥(𝑥 = 𝐴𝜑))
Colors of variables: wff set class
Syntax hints:  wi 4  wa 104  wb 105  wal 1371   = wceq 1373  wex 1516  wcel 2177  Vcvv 2773
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1471  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-11 1530  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-ext 2188
This theorem depends on definitions:  df-bi 117  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-v 2775
This theorem is referenced by:  ceqex  2902
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