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Mirrors > Home > ILE Home > Th. List > sbor | GIF version |
Description: Logical OR inside and outside of substitution are equivalent. (Contributed by NM, 29-Sep-2002.) (Proof rewritten by Jim Kingdon, 3-Feb-2018.) |
Ref | Expression |
---|---|
sbor | ⊢ ([𝑦 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∨ [𝑦 / 𝑥]𝜓)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sborv 1890 | . . . 4 ⊢ ([𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓)) | |
2 | 1 | sbbii 1765 | . . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓)) |
3 | sborv 1890 | . . 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 1526 | . . 3 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑧(𝜑 ∨ 𝜓)) | |
6 | 5 | sbco2vh 1945 | . 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑥](𝜑 ∨ 𝜓)) |
7 | ax-17 1526 | . . . 4 ⊢ (𝜑 → ∀𝑧𝜑) | |
8 | 7 | sbco2vh 1945 | . . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑) |
9 | ax-17 1526 | . . . 4 ⊢ (𝜓 → ∀𝑧𝜓) | |
10 | 9 | sbco2vh 1945 | . . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓) |
11 | 8, 10 | orbi12i 764 | . 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)) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 105 ∨ wo 708 [wsb 1762 |
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-4 1510 ax-17 1526 ax-i9 1530 ax-ial 1534 ax-i5r 1535 |
This theorem depends on definitions: df-bi 117 df-nf 1461 df-sb 1763 |
This theorem is referenced by: sbcor 3008 sbcorg 3009 unab 3403 |
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