Theorem sbc6g 3659

Description: An equivalence for class substitution. (Contributed by NM, 11-Oct-2004.) (Proof shortened by Andrew Salmon, 8-Jun-2011.)

Assertion

Ref	Expression
sbc6g	⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))

Distinct variable group:   𝑥,𝐴

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

Proof of Theorem sbc6g

Step	Hyp	Ref	Expression
1		alexeqg 3521	. 2 ⊢ (𝐴 ∈ 𝑉 → (∀𝑥(𝑥 = 𝐴 → 𝜑) ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑)))
2		sbc5 3658	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ ∃𝑥(𝑥 = 𝐴 ∧ 𝜑))
3	1, 2	syl6rbbr 282	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))

Syntax hints:   → wi 4   ↔ wb 198   ∧ wa 385  ∀wal 1651   = wceq 1653  ∃wex 1875   ∈ wcel 2157  [wsbc 3633

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1891  ax-4 1905  ax-5 2006  ax-6 2072  ax-7 2107  ax-9 2166  ax-10 2185  ax-12 2213  ax-ext 2777

This theorem depends on definitions:  df-bi 199  df-an 386  df-or 875  df-tru 1657  df-ex 1876  df-nf 1880  df-sb 2065  df-clab 2786  df-cleq 2792  df-clel 2795  df-v 3387  df-sbc 3634

This theorem is referenced by:  sbc6  3660  sbciegft  3664  ralsnsg  4407  fz1sbc  12670  rdgeqoa  33716  pm14.122a  39404