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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 1862 | . . . 4 ⊢ ([𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓)) | |
2 | 1 | sbbii 1738 | . . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∨ [𝑧 / 𝑥]𝜓)) |
3 | sborv 1862 | . . 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 1506 | . . 3 ⊢ ((𝜑 ∨ 𝜓) → ∀𝑧(𝜑 ∨ 𝜓)) | |
6 | 5 | sbco2vh 1916 | . 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∨ 𝜓) ↔ [𝑦 / 𝑥](𝜑 ∨ 𝜓)) |
7 | ax-17 1506 | . . . 4 ⊢ (𝜑 → ∀𝑧𝜑) | |
8 | 7 | sbco2vh 1916 | . . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑) |
9 | ax-17 1506 | . . . 4 ⊢ (𝜓 → ∀𝑧𝜓) | |
10 | 9 | sbco2vh 1916 | . . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓) |
11 | 8, 10 | orbi12i 753 | . 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)) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 104 ∨ wo 697 [wsb 1735 |
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 698 ax-5 1423 ax-7 1424 ax-gen 1425 ax-ie1 1469 ax-ie2 1470 ax-8 1482 ax-10 1483 ax-11 1484 ax-i12 1485 ax-4 1487 ax-17 1506 ax-i9 1510 ax-ial 1514 ax-i5r 1515 |
This theorem depends on definitions: df-bi 116 df-nf 1437 df-sb 1736 |
This theorem is referenced by: sbcor 2948 sbcorg 2949 unab 3338 |
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