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Proof step 2 from proof ls-at-subarray-pebs (previous step)

n1 < n0, n0 ≤ # apebs ⊦ subarray(apebs, 0, n0)[n1] = apebs[n1]

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)% ⊦ n0 < n → subarray(apebs, m, n)[n0] = apebs[m + n0]; (s)