Theorem sbcor 3732

Description: Distribution of class substitution over disjunction. (Contributed by NM, 31-Dec-2016.) (Revised by NM, 17-Aug-2018.)

Assertion

Ref	Expression
sbcor	⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))

Proof of Theorem sbcor

Dummy variable 𝑦 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sbcex 3691	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) → 𝐴 ∈ V)
2		sbcex 3691	. . 3 ⊢ ([𝐴 / 𝑥]𝜑 → 𝐴 ∈ V)
3		sbcex 3691	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
4	2, 3	jaoi 856	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
5		dfsbcq2 3684	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∨ 𝜓)))
6		dfsbcq2 3684	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
7		dfsbcq2 3684	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
8	6, 7	orbi12d 918	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
9		sbor 2312	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
10	5, 8, 9	vtoclbg 3473	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
11	1, 4, 10	pm5.21nii 383	1 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 209   ∨ wo 846   = wceq 1542  [wsb 2074   ∈ wcel 2114  Vcvv 3399  [wsbc 3681

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1802  ax-4 1816  ax-5 1917  ax-6 1975  ax-7 2020  ax-8 2116  ax-9 2124  ax-10 2145  ax-12 2179  ax-ext 2711

This theorem depends on definitions:  df-bi 210  df-an 400  df-or 847  df-tru 1545  df-ex 1787  df-nf 1791  df-sb 2075  df-clab 2718  df-cleq 2731  df-clel 2812  df-v 3401  df-sbc 3682

This theorem is referenced by:  sbcori  35913  sbc3or  41713  sbc3orgVD  42032  sbcoreleleqVD  42040