Theorem sbcor 3044

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 3008	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) → 𝐴 ∈ V)
2		sbcex 3008	. . 3 ⊢ ([𝐴 / 𝑥]𝜑 → 𝐴 ∈ V)
3		sbcex 3008	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
4	2, 3	jaoi 718	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
5		dfsbcq2 3002	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∨ 𝜓)))
6		dfsbcq2 3002	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
7		dfsbcq2 3002	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
8	6, 7	orbi12d 795	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
9		sbor 1983	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
10	5, 8, 9	vtoclbg 2835	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
11	1, 4, 10	pm5.21nii 706	1 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 105   ∨ wo 710   = wceq 1373  [wsb 1786   ∈ wcel 2177  Vcvv 2773  [wsbc 2999

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-io 711  ax-5 1471  ax-7 1472  ax-gen 1473  ax-ie1 1517  ax-ie2 1518  ax-8 1528  ax-10 1529  ax-11 1530  ax-i12 1531  ax-bndl 1533  ax-4 1534  ax-17 1550  ax-i9 1554  ax-ial 1558  ax-i5r 1559  ax-ext 2188

This theorem depends on definitions:  df-bi 117  df-tru 1376  df-nf 1485  df-sb 1787  df-clab 2193  df-cleq 2199  df-clel 2202  df-nfc 2338  df-v 2775  df-sbc 3000

This theorem is referenced by:  rabrsndc  3702