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Proof step 5 from proof sorted-inc-left (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
n < n0, n0 + 1 < # ar, n0 + 1 < m, ar[n0 + 1] < ar[n0], m ≤ # ar, ∀ n0. n < n0 ∧ n0 + 1 < m → ¬ ar[n0 + 1] < ar[n0] ⊦ n0 = n, ar[n + 1] < a

Rule argument

Additional information

Used simplifier rules: ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)