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Proof step 1 from proof ordered-shorter-sublist (conclusion)

⊦ ordered≤(sublist(m0, n1 + 1, x)) ∧ n2 < n1 ∧ n1 + m0 < # x → (¬ ordered≤(sublist(m0, n2 + 1, x)) ↔ false)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n1 + m0 < # x, n2 < n1, ordered≤(sublist(m0, n1 + 1, x)) ⊦ n1 = 0, x = [], ordered≤(sublist(m0, n2 + 1, x))

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1