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Mirrors > Home > ILE Home > Th. List > 3orcomb | GIF version |
Description: Commutation law for triple disjunction. (Contributed by Scott Fenton, 20-Apr-2011.) |
Ref | Expression |
---|---|
3orcomb | ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | orcom 728 | . . 3 ⊢ ((𝜓 ∨ 𝜒) ↔ (𝜒 ∨ 𝜓)) | |
2 | 1 | orbi2i 762 | . 2 ⊢ ((𝜑 ∨ (𝜓 ∨ 𝜒)) ↔ (𝜑 ∨ (𝜒 ∨ 𝜓))) |
3 | 3orass 981 | . 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ (𝜓 ∨ 𝜒))) | |
4 | 3orass 981 | . 2 ⊢ ((𝜑 ∨ 𝜒 ∨ 𝜓) ↔ (𝜑 ∨ (𝜒 ∨ 𝜓))) | |
5 | 2, 3, 4 | 3bitr4i 212 | 1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓)) |
Colors of variables: wff set class |
Syntax hints: ↔ wb 105 ∨ wo 708 ∨ w3o 977 |
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 709 |
This theorem depends on definitions: df-bi 117 df-3or 979 |
This theorem is referenced by: eueq3dc 2912 sotritrieq 4326 exmidontriimlem3 7222 |
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