Theorem sbc6gOLD 3808

Description: Obsolete version of sbc6g 3807 as of 5-Oct-2024. (Contributed by NM, 11-Oct-2004.) (Proof shortened by Andrew Salmon, 8-Jun-2011.) (Proof modification is discouraged.) (New usage is discouraged.)

Assertion

Ref	Expression
sbc6gOLD	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))

Distinct variable group:   𝑥,𝐴

Allowed substitution hints:   𝜑(𝑥)   𝑉(𝑥)

Proof of Theorem sbc6gOLD

Step	Hyp	Ref	Expression
1		sbc5 3805	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))
2		alexeqg 3639	. 2 ⊢ (𝐴 ∈ 𝑉 → (∀𝑥(𝑥 = 𝐴 → 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
3	1, 2	bitr4id 289	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396  ∀wal 1539   = wceq 1541  ∃wex 1781   ∈ wcel 2106  [wsbc 3777

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1797  ax-4 1811  ax-5 1913  ax-6 1971  ax-7 2011  ax-8 2108  ax-9 2116  ax-10 2137  ax-12 2171  ax-ext 2703

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 846  df-tru 1544  df-ex 1782  df-nf 1786  df-sb 2068  df-clab 2710  df-cleq 2724  df-clel 2810  df-sbc 3778

This theorem is referenced by: (None)