Theorem sbor 1928

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 1863	. . . 4 ⊢ ([𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓))
2	1	sbbii 1739	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓))
3		sborv 1863	. . 3 ⊢ ([𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∨ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
4	2, 3	bitri 183	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∨ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓))
5		ax-17 1507	. . 3 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑧(𝜑 ∨ 𝜓))
6	5	sbco2vh 1919	. 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑥](𝜑 ∨ 𝜓))
7		ax-17 1507	. . . 4 ⊢ (𝜑 → ∀𝑧𝜑)
8	7	sbco2vh 1919	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑)
9		ax-17 1507	. . . 4 ⊢ (𝜓 → ∀𝑧𝜓)
10	9	sbco2vh 1919	. . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓)
11	8, 10	orbi12i 754	. 2 ⊢ (([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ∨ [𝑦 / 𝑧][𝑧 / 𝑥]𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))
12	4, 6, 11	3bitr3i 209	1 ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓))

Syntax hints:   ↔ wb 104   ∨ wo 698  [wsb 1736

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 105  ax-ia2 106  ax-ia3 107  ax-io 699  ax-5 1424  ax-7 1425  ax-gen 1426  ax-ie1 1470  ax-ie2 1471  ax-8 1483  ax-10 1484  ax-11 1485  ax-i12 1486  ax-4 1488  ax-17 1507  ax-i9 1511  ax-ial 1515  ax-i5r 1516

This theorem depends on definitions:  df-bi 116  df-nf 1438  df-sb 1737

This theorem is referenced by:  sbcor  2957  sbcorg  2958  unab  3348