Theorem sbor 1876

Description: Logical OR inside and outside of substitution are equivalent. (Contributed by NM, 29-Sep-2002.) (Proof rewritten by Jim Kingdon, 3-Feb-2018.)

Assertion

Ref	Expression
sbor	⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Proof of Theorem sbor

Dummy variable 𝑧 is distinct from all other variables.

Step	Hyp	Ref	Expression
1		sborv 1818	. . . 4 ⊢ ([𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓))
2	1	sbbii 1695	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓))
3		sborv 1818	. . 3 ⊢ ([𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∨ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
4	2, 3	bitri 182	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∨ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
5		ax-17 1464	. . 3 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑧(𝜑 ∨ 𝜓))
6	5	sbco2v 1869	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑥](𝜑 ∨ 𝜓))
7		ax-17 1464	. . . 4 ⊢ (𝜑 → ∀𝑧𝜑)
8	7	sbco2v 1869	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
9		ax-17 1464	. . . 4 ⊢ (𝜓 → ∀𝑧𝜓)
10	9	sbco2v 1869	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓)
11	8, 10	orbi12i 716	. 2 ⊢ (([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∨ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
12	4, 6, 11	3bitr3i 208	1 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Syntax hints:   ↔ wb 103   ∨ wo 664  [wsb 1692

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-io 665  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-10 1441  ax-11 1442  ax-i12 1443  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473

This theorem depends on definitions:  df-bi 115  df-nf 1395  df-sb 1693

This theorem is referenced by:  sbcor  2883  sbcorg  2884  unab  3266