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Proof step 6 from proof ins-max (previous step)

¬ ordered≤(a ' + a0 ' + x), ordered≤(a0 ' + x), ∀ a1. ins≤(a1, x) = x + a1 ' ∨ a1 < max(x) ⊦ ins≤(a, a0 ' + x) = (a0 ' + x) + a ', a < max(a0 ' + x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information