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Proof step 6 from proof sorted-inc-right (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n < m0, ar[n0 + 1] < ar[n0], m0 + 1 < # ar, n0 + 1 < # ar, n + 1 < # ar, n < # ar, n0 < # ar, n0 < m0, n ≤ n0, ∀ n1. n ≤ n1 ∧ n1 < m0 → ¬ ar[n1 + 1] < ar[n1] ⊦ n0 = m0, a < ar[m0]

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → m ≤ n; (s)% ⊦ m < n → n ≠ 0; (s)%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic%m ≠ n ⊦ ar[m, a][n] = ar[n]; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)