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Proof step 3 from proof sorted-merge (previous step)

n0 < m, A[n1 + 1] < A[n1], n1 + 1 < m0, n0 < # A, n1 + 1 < # A, m0 ≤ # A, m ≤ # A, n ≤ n1, ∀ n1. n0 ≤ n1 ∧ n1 + 1 < m0 → ¬ A[n1 + 1] < A[n1], ∀ n1. n ≤ n1 ∧ n1 + 1 < m → ¬ A[n1 + 1] < A[n1] ⊦

Heuristic quantifier applied rule all left

And obtained the following new goals

Proof step 4:
n1 + 1 < # A, n0 < # A, n1 + 1 < m0, A[n1 + 1] < A[n1], n0 < m, n ≤ n1, m ≤ # A, m0 ≤ # A, n0 ≤ n1 ∧ n1 + 1 < m0 → ¬ A[n1 + 1] < A[n1], ∀ n1. n ≤ n1 ∧ n1 + 1 < m → ¬ A[n1 + 1] < A[n1], ∀ n1. n0 ≤ n1 ∧ n1 + 1 < m0 → ¬ A[n1 + 1] < A[n1] ⊦

Rule argument

pos: left 9 terms: n1

Additional information

Used terms: n1