Theorem sbceqbidf 28705

Description: Equality theorem for class substitution. (Contributed by Thierry Arnoux, 4-Sep-2018.)

Hypotheses

Ref	Expression
sbceqbidf.1	⊢ Ⅎ𝑥𝜑
sbceqbidf.2	⊢ (𝜑 → 𝐴 = 𝐵)
sbceqbidf.3	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
sbceqbidf	⊢ (𝜑 → ([𝐴 / 𝑥]𝜓 ↔ [𝐵 / 𝑥]𝜒))

Proof of Theorem sbceqbidf

Step	Hyp	Ref	Expression
1		sbceqbidf.2	. . 3 ⊢ (𝜑 → 𝐴 = 𝐵)
2		sbceqbidf.1	. . . 4 ⊢ Ⅎ𝑥𝜑
3		sbceqbidf.3	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
4	2, 3	abbid 2727	. . 3 ⊢ (𝜑 → {𝑥 ∣ 𝜓} = {𝑥 ∣ 𝜒})
5	1, 4	eleq12d 2682	. 2 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝐵 ∈ {𝑥 ∣ 𝜒}))
6		df-sbc 3403	. 2 ⊢ ([𝐴 / 𝑥]𝜓 ↔ 𝐴 ∈ {𝑥 ∣ 𝜓})
7		df-sbc 3403	. 2 ⊢ ([𝐵 / 𝑥]𝜒 ↔ 𝐵 ∈ {𝑥 ∣ 𝜒})
8	5, 6, 7	3bitr4g 302	1 ⊢ (𝜑 → ([𝐴 / 𝑥]𝜓 ↔ [𝐵 / 𝑥]𝜒))

Syntax hints:   → wi 4   ↔ wb 195   = wceq 1475  Ⅎwnf 1699   ∈ wcel 1977  {cab 2596  [wsbc 3402

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1713  ax-4 1728  ax-5 1827  ax-6 1875  ax-7 1922  ax-10 2006  ax-11 2021  ax-12 2034  ax-13 2234  ax-ext 2590

This theorem depends on definitions:  df-bi 196  df-or 384  df-an 385  df-tru 1478  df-ex 1696  df-nf 1701  df-sb 1868  df-clab 2597  df-cleq 2603  df-clel 2606  df-sbc 3403

This theorem is referenced by: (None)