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Proof step 14 from proof subarray-append-consec (previous step)

n < n0 + n1, n0 ≤ n, m0 + n0 + n1 ≤ # ar ⊦ subarray(ar, n0 + m0, n1)[n - n0] = ar[m0 + n], n1 = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, n0, n2]

Additional information

Used simplifier rules: ⊦ m ≤ m + n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, n0, n2