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Proof step 3 from proof sorted-inc-right (previous step)

n0 + 1 < # ar, m < # ar, ar[m, a][n0 + 1] < ar[n0], n0 < m, n < m, n0 < # ar, n < # ar, n + 1 < # ar, n ≤ n0, ∀ n0. n ≤ n0 ∧ n0 + 1 < m → ¬ ar[n0 + 1] < ar[n0] ⊦ a < ar[m - 1]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
m = m0 + 1, n + 1 < # ar, n < # ar, n0 < # ar, n < m, n0 < m, ar[m, a][n0 + 1] < ar[n0], m < # ar, n0 + 1 < # ar, n ≤ n0, ∀ n0. n ≤ n0 ∧ n0 + 1 < m → ¬ ar[n0 + 1] < ar[n0] ⊦ m = 0, a < ar[m0]

Rule argument

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

Additional information

Used simplifier rules: ⊦ 0 < m ↔ m ≠ 0; (s)% ⊦ m < n → n ≠ 0; (s)% ⊦ 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 ≠ n ⊦ ar[m, a][n] = ar[n]; (s)% ⊦ ¬ n < 0; (s)%n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1; (elim) Used terms: m, m0