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Mirrors > Home > MPE Home > Th. List > sb3an | Structured version Visualization version GIF version |
Description: Threefold 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 | sban 2083 | . . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓)) | |
2 | 1 | anbi1i 624 | . 2 ⊢ (([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒)) |
3 | df-3an 1088 | . . . 4 ⊢ ((𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ((𝜑 ∧ 𝜓) ∧ 𝜒)) | |
4 | 3 | sbbii 2079 | . . 3 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ [𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒)) |
5 | sban 2083 | . . 3 ⊢ ([𝑦 / 𝑥]((𝜑 ∧ 𝜓) ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒)) | |
6 | 4, 5 | bitri 274 | . 2 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝑦 / 𝑥](𝜑 ∧ 𝜓) ∧ [𝑦 / 𝑥]𝜒)) |
7 | df-3an 1088 | . 2 ⊢ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒) ↔ (([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓) ∧ [𝑦 / 𝑥]𝜒)) | |
8 | 2, 6, 7 | 3bitr4i 303 | 1 ⊢ ([𝑦 / 𝑥](𝜑 ∧ 𝜓 ∧ 𝜒) ↔ ([𝑦 / 𝑥]𝜑 ∧ [𝑦 / 𝑥]𝜓 ∧ [𝑦 / 𝑥]𝜒)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 205 ∧ wa 396 ∧ w3a 1086 [wsb 2067 |
This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1798 ax-4 1812 |
This theorem depends on definitions: df-bi 206 df-an 397 df-3an 1088 df-sb 2068 |
This theorem is referenced by: (None) |
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