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Proof step 1 from proof or-ass (conclusion)
⊦ ((P0 ∨ P1) ∨ P2) = (P0 ∨ P1 ∨ P2)
User applied rule expand right
And closed the goal!
Rule argument
pos: right 1
Additional information
Used simplifier rules: or: ⊦ ((P0 ∨ P1))(H) ↔ P0(H) ∨ P1(H);