Theorem sbcor 3009

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 2973	. 2 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) → 𝐴 ∈ V)
2		sbcex 2973	. . 3 ⊢ ([𝐴 / 𝑥]𝜑 → 𝐴 ∈ V)
3		sbcex 2973	. . 3 ⊢ ([𝐴 / 𝑥]𝜓 → 𝐴 ∈ V)
4	2, 3	jaoi 716	. 2 ⊢ (([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓) → 𝐴 ∈ V)
5		dfsbcq2 2967	. . 3 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝐴 / 𝑥](𝜑 ∨ 𝜓)))
6		dfsbcq2 2967	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜑 ↔ [𝐴 / 𝑥]𝜑))
7		dfsbcq2 2967	. . . 4 ⊢ (𝑦 = 𝐴 → ([𝑦 / 𝑥]𝜓 ↔ [𝐴 / 𝑥]𝜓))
8	6, 7	orbi12d 793	. . 3 ⊢ (𝑦 = 𝐴 → (([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
9		sbor 1954	. . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
10	5, 8, 9	vtoclbg 2800	. 2 ⊢ (𝐴 ∈ V → ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓)))
11	1, 4, 10	pm5.21nii 704	1 ⊢ ([𝐴 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝐴 / 𝑥]𝜑 ∨ [𝐴 / 𝑥]𝜓))

Syntax hints:   ↔ wb 105   ∨ wo 708   = wceq 1353  [wsb 1762   ∈ wcel 2148  Vcvv 2739  [wsbc 2964

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 709  ax-5 1447  ax-7 1448  ax-gen 1449  ax-ie1 1493  ax-ie2 1494  ax-8 1504  ax-10 1505  ax-11 1506  ax-i12 1507  ax-bndl 1509  ax-4 1510  ax-17 1526  ax-i9 1530  ax-ial 1534  ax-i5r 1535  ax-ext 2159

This theorem depends on definitions:  df-bi 117  df-tru 1356  df-nf 1461  df-sb 1763  df-clab 2164  df-cleq 2170  df-clel 2173  df-nfc 2308  df-v 2741  df-sbc 2965

This theorem is referenced by:  rabrsndc  3662