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Proof step 3 from proof sorted-lower-higher-bound (previous step)

A[n0 + 1] < A[n0], n0 + 1 < m, n0 + 1 < # A, n0 + 1 < m0, m ≤ # A, m0 ≤ # A, m ≤ m0, n ≤ n0, ∀ n0. n ≤ n0 ∧ n0 + 1 < m0 → ¬ A[n0 + 1] < A[n0] ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 9 terms: n0

Additional information

Used terms: n0