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Proof step 8 from proof valid-sqns-subarray-h (previous step)

n1 < # apebs, n1 < n, n ≤ # apebs, ∀ n, apebs, v, m. n ≤ # apebs ∧ n1 ≤ n ∧ n1 ≤ # apebs → valid-sqns(v, m, subarray(apebs, 0, n), n1) = valid-sqns(v, m, apebs, n1) ⊦ valid-sqns(v, m, subarray(apebs, 0, n), n1) = valid-sqns(v, m, apebs, n1)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: n, apebs, v, m

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s) Used terms: n, apebs, v, m