Theorem sbcbid 3800

Description: Formula-building deduction for class substitution. (Contributed by NM, 29-Dec-2014.)

Hypotheses

Ref	Expression
sbcbid.1	⊢ Ⅎ𝑥𝜑
sbcbid.2	⊢ (𝜑 → (𝜓 ↔ 𝜒))

Assertion

Ref	Expression
sbcbid	⊢ (𝜑 → ([𝐴 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜒))

Proof of Theorem sbcbid

Step	Hyp	Ref	Expression
1		sbcbid.1	. . . 4 ⊢ Ⅎ𝑥𝜑
2		sbcbid.2	. . . 4 ⊢ (𝜑 → (𝜓 ↔ 𝜒))
3	1, 2	abbid 2833	. . 3 ⊢ (𝜑 → {𝑥 ∣ 𝜓} = {𝑥 ∣ 𝜒})
4	3	eleq2d 2851	. 2 ⊢ (𝜑 → (𝐴 ∈ {𝑥 ∣ 𝜓} ↔ 𝐴 ∈ {𝑥 ∣ 𝜒}))
5		df-sbc 3747	. 2 ⊢ ([𝐴 / 𝑥]𝜓 ↔ 𝐴 ∈ {𝑥 ∣ 𝜓})
6		df-sbc 3747	. 2 ⊢ ([𝐴 / 𝑥]𝜒 ↔ 𝐴 ∈ {𝑥 ∣ 𝜒})
7	4, 5, 6	3bitr4g 317	1 ⊢ (𝜑 → ([𝐴 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜒))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209  Ⅎwnf 1816   ∈ 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-12 2216  ax-ext 2737

This proof depends on definitions:  df-bi 210  df-an 402  df-ex 1813  df-nf 1817  df-sb 2100  df-clab 2744  df-cleq 2757  df-clel 2840  df-sbc 3747

This theorem is used by:  csbeq2d  3860