Theorem sbceqbii 36980

Description: Formula-building inference for class substitution. General version of sbcbii 3794. (Contributed by GG, 1-Sep-2025.)

Hypotheses

Ref	Expression
sbceqbii.1	⊢ 𝐴 = 𝐵
sbceqbii.2	⊢ (𝜑 ↔ 𝜓)

Assertion

Ref	Expression
sbceqbii	⊢ ([𝐴 / 𝑥]𝜑 ↔ [𝐵 / 𝑥]𝜓)

Proof of Theorem sbceqbii

Step	Hyp	Ref	Expression
1		sbceqbii.1	. . 3 ⊢ 𝐴 = 𝐵
2		sbceqbii.2	. . . 4 ⊢ (𝜑 ↔ 𝜓)
3	2	abbii 2827	. . 3 ⊢ {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓}
4	1, 3	eleq12i 2853	. 2 ⊢ (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝐵 ∈ {𝑥 ∣ 𝜓})
5		df-sbc 3739	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
6		df-sbc 3739	. 2 ⊢ ([𝐵 / 𝑥]𝜓 ↔ 𝐵 ∈ {𝑥 ∣ 𝜓})
7	4, 5, 6	3bitr4i 306	1 ⊢ ([𝐴 / 𝑥]𝜑 ↔ [𝐵 / 𝑥]𝜓)

This proof depends on syntax axioms:   ↔ wb 209   = 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-ex 1812  df-sb 2099  df-clab 2739  df-cleq 2752  df-clel 2835  df-sbc 3739

This theorem is used by: (None)