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Proof step 19 from proof subarray-split-first (previous step)

n0 + m < # ar, 1 ≤ m, 1 ≤ n, n ≤ m ⊦ ar[n0 + n] = subarray(ar, n0 + 1, m)[n - 1], m = 0, n = 0

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

spec: nat name: elim-pred-c seq: n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1 subst: [n, m] → [n, m0]

Additional information

Used simplifier rules: ⊦ m + 1 ≠ 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ m + n; (s)% ⊦ m + 1 ≤ n ↔ m < n; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ n0 < n → subarray(ar, m, n)[n0] = ar[m + n0]; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: n, m0