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Proof step 4 from proof ordered-shorter-sublist (previous step)

sublist(m0, n2 + 1, x) ⊑ sublist(m0, n1 + 1, x) ∧ ordered≤(sublist(m0, n1 + 1, x)) → (¬ ¬ ordered≤(sublist(m0, n2 + 1, x)) ↔ true), n1 + m0 < # x, n2 < n1, ordered≤(sublist(m0, n1 + 1, x)) ⊦ n1 = 0, x = [], ordered≤(sublist(m0, n2 + 1, x))

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: left 1

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