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Proof step 7 from proof ordered-prefix (previous step)

a = x0.head, ordered≤(x0), x1 ⊑ x0.tail, (∃ x0. ordered≤(x0) ∧ x1 ⊑ x0) → ordered≤(x1) ⊦ x0 = [], ordered≤(a ' + x1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information