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Mirrors > Home > ILE Home > Th. List > sb3an | GIF version |
Description: Conjunction inside and outside of a substitution are equivalent. (Contributed by NM, 14-Dec-2006.) |
Ref | Expression |
---|---|
sb3an | ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | df-3an 965 | . . 3 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒)) | |
2 | 1 | sbbii 1739 | . 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒)) |
3 | sban 1929 | . 2 ⊢ ([𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒)) | |
4 | sban 1929 | . . . 4 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓)) | |
5 | 4 | anbi1i 454 | . . 3 ⊢ (([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒)) |
6 | df-3an 965 | . . 3 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒)) | |
7 | 5, 6 | bitr4i 186 | . 2 ⊢ (([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒)) |
8 | 2, 3, 7 | 3bitri 205 | 1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒)) |
Colors of variables: wff set class |
Syntax hints: ∧ wa 103 ↔ wb 104 ∧ w3a 963 [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-3an 965 df-nf 1438 df-sb 1737 |
This theorem is referenced by: (None) |
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