Theorem sbceqbidf 32844

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 2830	. . 3 ⊢ (𝜑 → {𝑥 ∣ 𝜓} = {𝑥 ∣ 𝜒})
5	1, 4	eleq12d 2856	. 2 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝐵 ∈ {𝑥 ∣ 𝜒}))
6		df-sbc 3744	. 2 ⊢ ([𝐴 / 𝑥]𝜓 ↔ 𝐴 ∈ {𝑥 ∣ 𝜓})
7		df-sbc 3744	. 2 ⊢ ([𝐵 / 𝑥]𝜒 ↔ 𝐵 ∈ {𝑥 ∣ 𝜒})
8	5, 6, 7	3bitr4g 317	1 ⊢ (𝜑 → ([𝐴 / 𝑥]𝜓 ↔ [𝐵 / 𝑥]𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209   = wceq 1569  Ⅎwnf 1812   ∈ wcel 2142  {cab 2740  [wsbc 3743

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1824  ax-4 1838  ax-5 1939  ax-6 1996  ax-7 2037  ax-8 2144  ax-9 2152  ax-12 2212  ax-ext 2734

This proof depends on definitions:  df-bi 210  df-an 401  df-ex 1809  df-nf 1813  df-sb 2096  df-clab 2741  df-cleq 2754  df-clel 2837  df-sbc 3744

This theorem is used by: (None)