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Proof step 4 from proof sorted-insert-merge (previous step)

m = m0 + 1, n0 + 1 < # A, n0 + 1 < n, A[m, a][n0 + 1] < A[m, a][n0], m ≤ # A, n ≤ # A, ∀ n0. n0 + 1 < m → ¬ A[n0 + 1] < A[n0], ∀ n0. m < n0 ∧ n0 + 1 < n → ¬ A[n0 + 1] < A[n0] ⊦ m = 0, A[m + 1] < A[m0], a < A[m0], A[m + 1] < a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
A[m0 + 1, a][n0 + 1] < A[m0 + 1, a][n0], n0 + 1 < n, n0 + 1 < # A, m0 < # A, n ≤ # A, ∀ n0. m0 + 1 < n0 ∧ n0 + 1 < n → ¬ A[n0 + 1] < A[n0], ∀ n1. n1 < m0 → ¬ A[n1 + 1] < A[n1] ⊦ A[2 + m0] < a, a < A[m0], A[2 + m0] < A[m0]

Rule argument

Additional information

Used simplifier rules: nnn: ⊦ ¬ c < b ∧ ¬ a < c → ¬ a < b; (forward) from specification oelem% ⊦ m + n < m + n0 ↔ n < n0; (s)% ⊦ m + 1 ≤ 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 + 1 ≠ 0; (s)