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Proof step 23 from proof append-a (previous step)

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

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, # ar, n0]

Additional information

Used simplifier rules: append-len: ⊦ #(ar0 + ar1) = # ar0 + # ar1;%append-atp2: n < # ar1 ⊦ (ar0 + ar1)[# ar0 + n] = ar1[n];%append-at1: ⊦ n < # ar0 → (ar0 + ar1)[n] = ar0[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 < m0 → m + n < m0 + n0; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat% ⊦ ¬ m + n < m; (s)%n ≤ m ⊦ n0 = m - n ↔ m = n0 + n; (elim) Used terms: n, # ar, n0