Metamath Proof Explorer |
< Previous
Next >
Nearby theorems |
||
Mirrors > Home > MPE Home > Th. List > 3orcomb | Structured version Visualization version GIF version |
Description: Commutation law for triple disjunction. (Contributed by Scott Fenton, 20-Apr-2011.) (Proof shortened by Wolf Lammen, 8-Apr-2022.) |
Ref | Expression |
---|---|
3orcomb | ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓)) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | 3orcoma 1090 | . 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒)) | |
2 | 3orrot 1089 | . 2 ⊢ ((𝜓 ∨ 𝜑 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓)) | |
3 | 1, 2 | bitri 278 | 1 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜑 ∨ 𝜒 ∨ 𝜓)) |
Colors of variables: wff setvar class |
Syntax hints: ↔ wb 209 ∨ w3o 1083 |
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 210 df-or 845 df-3or 1085 |
This theorem is referenced by: eueq3 3627 swoso 8337 swrdnd 14068 colcom 26456 legso 26497 lncom 26520 soseq 33362 colinearperm1 33939 frege129d 40865 ordelordALT 41644 ordelordALTVD 41974 |
Copyright terms: Public domain | W3C validator |