Theorem sbcor 3794

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 3754	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) → 𝐴 ∈ V)
2		sbcex 3754	. . 3 ⊢ ([𝐴 / 𝑥]𝜑 → 𝐴 ∈ V)
3		sbcex 3754	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
4	2, 3	jaoi 868	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
5		dfsbcq2 3747	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∨ 𝜓)))
6		dfsbcq2 3747	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
7		dfsbcq2 3747	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
8	6, 7	orbi12d 929	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
9		sbor 2339	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
10	5, 8, 9	vtoclbg 3523	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
11	1, 4, 10	pm5.21nii 380	1 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 208   ∨ wo 858   = wceq 1559  [wsb 2089   ∈ wcel 2141  Vcvv 3453  [wsbc 3744

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8  ax-gen 1814  ax-4 1828  ax-5 1929  ax-6 1986  ax-7 2027  ax-8 2143  ax-9 2151  ax-10 2174  ax-12 2211  ax-ext 2733

This theorem depends on definitions:  df-bi 209  df-an 400  df-or 859  df-tru 1562  df-ex 1799  df-nf 1803  df-sb 2090  df-clab 2740  df-cleq 2753  df-clel 2836  df-v 3455  df-sbc 3745

This theorem is referenced by:  sbcori  38572  sbc3or  45072  sbc3orgVD  45390  sbcoreleleqVD  45398