Theorem sbcbi2 3802

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 2176, ax-12 2213. (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 2828	. . 3 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓})
2	1	eleq2d 2849	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜓}))
3		df-sbc 3745	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
4		df-sbc 3745	. 2 ⊢ ([𝐴 / 𝑥]𝜓 ↔ 𝐴 ∈ {𝑥 ∣ 𝜓})
5	2, 3, 4	3bitr4g 317	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))

Syntax hints:   → wi 4   ↔ wb 209  ∀wal 1568   ∈ wcel 2143  {cab 2741  [wsbc 3744

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1825  ax-4 1839  ax-5 1940  ax-6 1997  ax-7 2038  ax-8 2145  ax-9 2153  ax-ext 2735

This theorem depends on definitions:  df-bi 210  df-an 401  df-ex 1810  df-sb 2097  df-clab 2742  df-cleq 2755  df-clel 2838  df-sbc 3745

This theorem is referenced by:  csbeq2  3858