Theorem sbc6g 3768

Description: An equivalence for class substitution. (Contributed by NM, 11-Oct-2004.) (Proof shortened by Andrew Salmon, 8-Jun-2011.) (Proof shortened by SN, 5-Oct-2024.)

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		df-sbc 3739	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
2		elab6g 3622	. 2 ⊢ (𝐴 ∈ 𝑉 → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))
3	1, 2	bitrid 286	1 ⊢ (𝐴 ∈ 𝑉 → ([𝐴 / 𝑥]𝜑 ↔ ∀𝑥(𝑥 = 𝐴 → 𝜑)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1567   = wceq 1569   ∈ wcel 2144  {cab 2738  [wsbc 3738

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1827  ax-4 1841  ax-5 1942  ax-6 1999  ax-7 2040  ax-8 2146  ax-9 2154  ax-ext 2732

This proof depends on definitions:  df-bi 210  df-an 402  df-tru 1572  df-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-sbc 3739

This theorem is used by:  sbc6  3769  sbciegft  3775  sbccomlem  3816  ralsnsg  4627  fz1sbc  13737  rdgeqoa  38386  pm14.122a  45499