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Proof step 25 from proof copy-shift-right (previous step)

n < # ar0 + # ar1, # ar0 ≤ n, n < n0 + # ar0 ∨ (n0 + # ar0) + m1 ≤ n ⊦ # ar1 = 0, ar1[n - # ar0] = copy(ar, m, ar1, n0, m1)[n - # ar0]

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, # ar0, n1]

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)%⊦ m0 < # ar1 → copy(ar0, n, ar1, n0, m)[m0] = (m0 < n0 ∨ n0 + m ≤ m0 ⊃ ar1[m0];ar0[m0 + n - n0]); (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # ar0, n1