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| Mirrors > Home > MPE Home > Th. List > Mathboxes > 3o3cs | Structured version Visualization version GIF version | ||
| Description: Deduction eliminating disjunct. (Contributed by Thierry Arnoux, 19-Dec-2016.) | 
| Ref | Expression | 
|---|---|
| 3o1cs.1 | ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) → 𝜃) | 
| Ref | Expression | 
|---|---|
| 3o3cs | ⊢ (𝜒 → 𝜃) | 
| Step | Hyp | Ref | Expression | 
|---|---|---|---|
| 1 | df-3or 1088 | . . 3 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒)) | |
| 2 | 3o1cs.1 | . . 3 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) → 𝜃) | |
| 3 | 1, 2 | sylbir 235 | . 2 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) → 𝜃) | 
| 4 | 3 | olcs 877 | 1 ⊢ (𝜒 → 𝜃) | 
| Colors of variables: wff setvar class | 
| Syntax hints: → wi 4 ∨ wo 848 ∨ w3o 1086 | 
| This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 | 
| This theorem depends on definitions: df-bi 207 df-or 849 df-3or 1088 | 
| This theorem is referenced by: xrpxdivcld 32917 | 
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