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Proof step 6 from proof ordered-prefix (previous step)

ordered≤(x0), x0 ≠ [] ∧ a = x0.first ∧ x1 ⊑ x0.rest, (∃ x0. x1 ⊑ x0 ∧ ordered≤(x0)) → ordered≤(x1) ⊦ ordered≤(a ' + x1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information