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Theorem ralsnsg 4632
Description: Substitution expressed in terms of quantification over a singleton. (Contributed by NM, 14-Dec-2005.) (Revised by Mario Carneiro, 23-Apr-2015.)
Assertion
Ref Expression
ralsnsg (𝐴 ∈ 𝑉 → (∀𝑥 ∈ {𝐴}𝜑 ↔ [𝐴 / 𝑥]𝜑))
Distinct variable group:   𝑥,𝐴
Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem ralsnsg
StepHypRef Expression
1 df-ral 3081 . . 3 (∀𝑥 ∈ {𝐴}𝜑 ↔ ∀𝑥(𝑥 ∈ {𝐴} → 𝜑))
2 velsn 4600 . . . . 5 (𝑥 ∈ {𝐴} ↔ 𝑥 = 𝐴)
32imbi1i 352 . . . 4 ((𝑥 ∈ {𝐴} → 𝜑) ↔ (𝑥 = 𝐴 → 𝜑))
43albii 1852 . . 3 (∀𝑥(𝑥 ∈ {𝐴} → 𝜑) ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑))
51, 4bitri 278 . 2 (∀𝑥 ∈ {𝐴}𝜑 ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑))
6 sbc6g 3772 . 2 (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))
75, 6bitr4id 293 1 (𝐴 ∈ 𝑉 → (∀𝑥 ∈ {𝐴}𝜑 ↔ [𝐴 / 𝑥]𝜑))
Colors of variables:    wff setvar class
This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568   = wceq 1570   ∈ wcel 2147  ∀wral 3080  [wsbc 3742  {csn 4588
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-ext 2735
This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1573  df-ex 1813  df-sb 2102  df-clab 2742  df-cleq 2755  df-clel 2839  df-ral 3081  df-v 3456  df-sbc 3743  df-sn 4589
This theorem is used by:  ralsngf  4635  ixpsnval  8966  ac6sfi  9313  rexfiuz  15559  prmind2  16904  finixpnum  38796
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