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Proof step 2 from proof subarray-subarray-elim-0 (previous step)

n1 < # ar0, n2 ≤ n1 ⊦ subarray(subarray(ar0, 0, n1), 0, n2) = subarray(ar0, 0, n2)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Subarray: ⊦ subarray(ar, m, n) = copy(ar, m, mkarray(n), 0, n);%subarray-copy-elim: m1 + n ≤ # ar1 ⊦ n1 ≤ n → subarray(copy(ar0, m0, ar1, m1, n), m1, n1) = subarray(ar0, m0, n1);% ⊦ ¬ n < n; (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ # mkarray(n) = n; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)