Theorem sbcor 2807

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

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 2772	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) → 𝐴 ∈ V)
2		sbcex 2772	. . 3 ⊢ ([𝐴 / 𝑥]𝜑 → 𝐴 ∈ V)
3		sbcex 2772	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
4	2, 3	jaoi 636	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
5		dfsbcq2 2767	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∨ 𝜓)))
6		dfsbcq2 2767	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
7		dfsbcq2 2767	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
8	6, 7	orbi12d 707	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
9		sbor 1828	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
10	5, 8, 9	vtoclbg 2614	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
11	1, 4, 10	pm5.21nii 620	1 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 98   ∨ wo 629   = wceq 1243   ∈ wcel 1393  [wsb 1645  Vcvv 2557  [wsbc 2764

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 99  ax-ia2 100  ax-ia3 101  ax-io 630  ax-5 1336  ax-7 1337  ax-gen 1338  ax-ie1 1382  ax-ie2 1383  ax-8 1395  ax-10 1396  ax-11 1397  ax-i12 1398  ax-bndl 1399  ax-4 1400  ax-17 1419  ax-i9 1423  ax-ial 1427  ax-i5r 1428  ax-ext 2022

This theorem depends on definitions:  df-bi 110  df-tru 1246  df-nf 1350  df-sb 1646  df-clab 2027  df-cleq 2033  df-clel 2036  df-nfc 2167  df-v 2559  df-sbc 2765

This theorem is referenced by:  rabrsndc  3438