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Proof step 6 from proof ordered-shorter-sublist (previous step)

n2 < n1, n1 + m0 < # x, ordered≤(sublist(m0, n1 + 1, x)) ⊦ x = [], ordered≤(sublist(m0, n2 + 1, x)), sublist(m0, n2 + 1, x) ⊑ sublist(m0, n1 + 1, x) ∧ ordered≤(sublist(m0, n1 + 1, x))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n1 + m0 < # x, n2 < n1, ordered≤(sublist(m0, n1 + 1, x)) ⊦ x = [], ordered≤(sublist(m0, n2 + 1, x)), sublist(m0, n2 + 1, x) ⊑ sublist(m0, n1 + 1, x)

Proof step 6 from proof ordered-shorter-sublist (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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