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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 1087 | . . 3 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ ((𝜑 ∨ 𝜓) ∨ 𝜒)) | |
2 | 3o1cs.1 | . . 3 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) → 𝜃) | |
3 | 1, 2 | sylbir 234 | . 2 ⊢ (((𝜑 ∨ 𝜓) ∨ 𝜒) → 𝜃) |
4 | 3 | olcs 873 | 1 ⊢ (𝜒 → 𝜃) |
Colors of variables: wff setvar class |
Syntax hints: → wi 4 ∨ wo 844 ∨ w3o 1085 |
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 206 df-or 845 df-3or 1087 |
This theorem is referenced by: xrpxdivcld 31209 |
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