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Proof step 4 from proof sorted-inc-right (previous step)

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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
n0 < # ar, n < # ar, n + 1 < # ar, n0 + 1 < # ar, m0 + 1 < # ar, ar[m0 + 1, a][n0 + 1] < ar[n0], n ≤ n0, n0 ≤ m0, n ≤ m0, ∀ n1. n ≤ n1 ∧ n1 < m0 → ¬ ar[n1 + 1] < ar[n1] ⊦ a < ar[m0]

Rule argument

Additional information

Used simplifier rules: let: ⊦ n ≤ n0 ∧ n0 ≤ n1 → n ≤ n1; (forward) from specification nat% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n + 1 ↔ ¬ n < m; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + 1 ≠ 0; (s)