Theorem sbc6gOLD 3809

Description: Obsolete version of sbc6g 3808 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 3806	. 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 394  ∀wal 1531   = wceq 1533  ∃wex 1773   ∈ wcel 2098  [wsbc 3778

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1789  ax-4 1803  ax-5 1905  ax-6 1963  ax-7 2003  ax-8 2100  ax-9 2108  ax-10 2129  ax-12 2166  ax-ext 2699

This theorem depends on definitions:  df-bi 206  df-an 395  df-or 846  df-tru 1536  df-ex 1774  df-nf 1778  df-sb 2060  df-clab 2706  df-cleq 2720  df-clel 2806  df-sbc 3779

This theorem is referenced by: (None)