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Proof step 2 from proof ordered-shorter-sublist (previous step)

n2 < n1, n1 + m0 < # x, ordered≤(sublist(m0, n1 + 1, x)) ⊦ ordered≤(sublist(m0, n2 + 1, x)) ↔ true

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n2 < n1, n1 + m0 < # x, ordered≤(sublist(m0, n1 + 1, x)) ⊦ x = [], ordered≤(sublist(m0, n2 + 1, x))

Rule argument

Additional information

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