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Proof step 3 from proof append-at2 (previous step)

n < # ar0 + # ar1, # ar0 ≤ n ⊦ # ar1 = 0, (ar0 + ar1)[n] = ar1[n - # ar0]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Rule argument

spec: nat name: elim-sub seq: n ≤ m ⊦ n0 = m - n ↔ m = n0 + n subst: [m, n, n0] → [n, # ar0, n0]

Additional information

Used simplifier rules: append-atp2: n < # ar1 ⊦ (ar0 + ar1)[# ar0 + n] = ar1[n];% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ 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 < m ↔ false); (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # ar0, n0