Theorem sbc6gOLD 3751

Description: Obsolete version of sbc6g 3750 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 3748	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))
2		alexeqg 3582	. 2 ⊢ (𝐴 ∈ 𝑉 → (∀𝑥(𝑥 = 𝐴 → 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
3	1, 2	bitr4id 290	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))

Syntax hints:   → wi 4   ↔ wb 205   ∧ wa 396  ∀wal 1540   = wceq 1542  ∃wex 1786   ∈ wcel 2110  [wsbc 3720

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2015  ax-8 2112  ax-9 2120  ax-10 2141  ax-12 2175  ax-ext 2711

This theorem depends on definitions:  df-bi 206  df-an 397  df-or 845  df-tru 1545  df-ex 1787  df-nf 1791  df-sb 2072  df-clab 2718  df-cleq 2732  df-clel 2818  df-sbc 3721

This theorem is referenced by: (None)