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Proof step 1 from proof max-ins-max-last (conclusion)

⊦ ¬ a < max(x0) ∧ ordered≤(x0) → ins≤(a, x0) = x0 + a '

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
ordered≤(x0) ⊦ ins≤(a, x0) = x0 + a ', a < max(x0)

Rule argument

Additional information