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Mirrors > Home > MPE Home > Th. List > 3orcoma | Structured version Visualization version GIF version |
Description: Commutation law for triple disjunction. (Contributed by Mario Carneiro, 4-Sep-2016.) |
Ref | Expression |
---|---|
3orcoma | ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | or12 915 | . 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ (𝜓 ∨ (𝜑 ∨ 𝜒))) | |
2 | 3orass 1083 | . 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒))) | |
3 | 3orass 1083 | . 2 ⊢ ((𝜓 ∨ 𝜑 ∨ 𝜒) ↔ (𝜓 ∨ (𝜑 ∨ 𝜒))) | |
4 | 1, 2, 3 | 3bitr4i 304 | 1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 207 ∨ wo 842 ∨ w3o 1079 |
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 208 df-or 843 df-3or 1081 |
This theorem is referenced by: 3orcomb 1087 outpasch 26227 eliccioo 30287 |
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