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Proof step 10 from proof sorted-insert-merge (previous step)

m0 < # A, n0 + 1 < # A, n0 + 1 < n, A[n0 + 1] < A[n0], n ≤ # A, ∀ n1. n1 < m0 → ¬ A[n1 + 1] < A[n1], ∀ n0. m0 + 1 < n0 ∧ n0 + 1 < n → ¬ A[n0 + 1] < A[n0] ⊦ n0 = m0, n0 = m0 + 1, A[2 + m0] < A[m0], a < A[m0], A[2 + m0] < a

User applied rule all left

And obtained the following new goals

Proof step 11:
A[n0 + 1] < A[n0], n0 + 1 < n, n0 + 1 < # A, m0 < # A, n ≤ # A, n0 < m0 → ¬ A[n0 + 1] < A[n0], ∀ n0. m0 + 1 < n0 ∧ n0 + 1 < n → ¬ A[n0 + 1] < A[n0], ∀ n1. n1 < m0 → ¬ A[n1 + 1] < A[n1] ⊦ n0 = m0 + 1, n0 = m0, A[2 + m0] < a, a < A[m0], A[2 + m0] < A[m0]

Rule argument

pos: left 6 terms: n0 abort: F discard: F computed: T

Additional information

Used terms: n0