Theorem sbcbi2 3804

Description: Substituting into equivalent wff's gives equivalent results. (Contributed by Giovanni Mascellani, 9-Apr-2018.) (Proof shortened by Wolf Lammen, 4-May-2023.) Avoid ax-10 2179, ax-12 2216. (Revised by Steven Nguyen, 5-May-2024.)

Assertion

Ref	Expression
sbcbi2	⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))

Proof of Theorem sbcbi2

Step	Hyp	Ref	Expression
1		abbi 2830	. . 3 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓})
2	1	eleq2d 2851	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜓}))
3		df-sbc 3747	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
4		df-sbc 3747	. 2 ⊢ ([𝐴 / 𝑥]𝜓 ↔ 𝐴 ∈ {𝑥 ∣ 𝜓})
5	2, 3, 4	3bitr4g 317	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  ∀wal 1568   ∈ wcel 2146  {cab 2743  [wsbc 3746

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1828  ax-4 1842  ax-5 1943  ax-6 2000  ax-7 2041  ax-8 2148  ax-9 2156  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-sbc 3747

This theorem is used by:  csbeq2  3859