Theorem sbor 1970

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 1902	. . . 4 ⊢ ([𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓))
2	1	sbbii 1776	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓))
3		sborv 1902	. . 3 ⊢ ([𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∨ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
4	2, 3	bitri 184	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∨ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
5		ax-17 1537	. . 3 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑧(𝜑 ∨ 𝜓))
6	5	sbco2vh 1961	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑥](𝜑 ∨ 𝜓))
7		ax-17 1537	. . . 4 ⊢ (𝜑 → ∀𝑧𝜑)
8	7	sbco2vh 1961	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
9		ax-17 1537	. . . 4 ⊢ (𝜓 → ∀𝑧𝜓)
10	9	sbco2vh 1961	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓)
11	8, 10	orbi12i 765	. 2 ⊢ (([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∨ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
12	4, 6, 11	3bitr3i 210	1 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Syntax hints:   ↔ wb 105   ∨ wo 709  [wsb 1773

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 710  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-10 1516  ax-11 1517  ax-i12 1518  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546

This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774

This theorem is referenced by:  sbcor  3030  sbcorg  3031  unab  3426