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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 1901 | . . . 4 ⊢ ([𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓)) | |
2 | 1 | sbbii 1776 | . . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝑦 / 𝑧]([𝑧 / 𝑥]𝜑 ∧ [𝑧 / 𝑥]𝜓)) |
3 | sbanv 1901 | . . 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 1537 | . . 3 ⊢ ((𝜑 ∧ 𝜓) → ∀𝑧(𝜑 ∧ 𝜓)) | |
6 | 5 | sbco2vh 1957 | . 2 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥](𝜑 ∧ 𝜓) ↔ [𝑦 / 𝑥](𝜑 ∧ 𝜓)) |
7 | ax-17 1537 | . . . 4 ⊢ (𝜑 → ∀𝑧𝜑) | |
8 | 7 | sbco2vh 1957 | . . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜑 ↔ [𝑦 / 𝑥]𝜑) |
9 | ax-17 1537 | . . . 4 ⊢ (𝜓 → ∀𝑧𝜓) | |
10 | 9 | sbco2vh 1957 | . . 3 ⊢ ([𝑦 / 𝑧][𝑧 / 𝑥]𝜓 ↔ [𝑦 / 𝑥]𝜓) |
11 | 8, 10 | anbi12i 460 | . 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: ∧ wa 104 ↔ wb 105 [wsb 1773 |
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 710 ax-5 1458 ax-7 1459 ax-gen 1460 ax-ie1 1504 ax-ie2 1505 ax-8 1515 ax-10 1516 ax-11 1517 ax-i12 1518 ax-4 1521 ax-17 1537 ax-i9 1541 ax-ial 1545 ax-i5r 1546 |
This theorem depends on definitions: df-bi 117 df-nf 1472 df-sb 1774 |
This theorem is referenced by: sb3an 1970 sbbi 1971 sbmo 2097 moanim 2112 sbabel 2359 nfrexdya 2526 cbvreu 2716 rmo3f 2949 sbcan 3020 sbcang 3021 rmo3 3069 inab 3418 difab 3419 exss 4245 inopab 4777 bdcriota 15093 |
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