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Proof step 18 from proof subarray-split-at (previous step)

# ar = n0 + n, n1 < n0 + n, n1 < n + n0, n ≤ n1 ⊦ ar[n1] = subarray(ar, n, n0)[n1 - n], n0 = 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] → [n1, n, 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: n1, n, n2