Theorem sbcbi2 3814

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 2142, ax-12 2178. (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 2795	. . 3 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → {𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓})
2		eleq2 2818	. . 3 ⊢ ({𝑥 ∣ 𝜑} = {𝑥 ∣ 𝜓} → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜓}))
3	1, 2	syl 17	. 2 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → (𝐴 ∈ {𝑥 ∣ 𝜑} ↔ 𝐴 ∈ {𝑥 ∣ 𝜓}))
4		df-sbc 3756	. 2 ⊢ ([𝐴 / 𝑥]𝜑 ↔ 𝐴 ∈ {𝑥 ∣ 𝜑})
5		df-sbc 3756	. 2 ⊢ ([𝐴 / 𝑥]𝜓 ↔ 𝐴 ∈ {𝑥 ∣ 𝜓})
6	3, 4, 5	3bitr4g 314	1 ⊢ (∀𝑥(𝜑 ↔ 𝜓) → ([𝐴 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜓))

Syntax hints:   → wi 4   ↔ wb 206  ∀wal 1538   = wceq 1540   ∈ wcel 2109  {cab 2708  [wsbc 3755

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1795  ax-4 1809  ax-5 1910  ax-6 1967  ax-7 2008  ax-8 2111  ax-9 2119  ax-ext 2702

This theorem depends on definitions:  df-bi 207  df-an 396  df-ex 1780  df-sb 2066  df-clab 2709  df-cleq 2722  df-clel 2804  df-sbc 3756

This theorem is referenced by:  csbeq2  3869