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Mirrors > Home > ILE Home > Th. List > sban | GIF version |
Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 5-Aug-1993.) (Proof rewritten by Jim Kingdon, 3-Feb-2018.) |
Ref | Expression |
---|---|
sban | ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | sbanv 1861 | . . . 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 | sbanv 1861 | . . 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 | anbi12i 455 | . 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: ∧ wa 103 ↔ wb 104 [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: sb3an 1929 sbbi 1930 sbmo 2056 moanim 2071 sbabel 2305 nfrexdya 2468 cbvreu 2650 rmo3f 2876 sbcan 2946 sbcang 2947 rmo3 2995 inab 3339 difab 3340 exss 4144 inopab 4666 bdcriota 13070 |
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