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Proof step 6 from proof sorted-shift-one (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m + 1 < # A, m + 1 < n0, 2 + m < n0, A[2 + m] < A[m], n < # A, n < n0, 2 + m < # A, n ≤ m + 1, n0 ≤ # A, n ≤ m, ∀ n1. n ≤ n1 ∧ n1 + 1 < n0 → ¬ A[n1 + 1] < A[n1] ⊦

Rule argument

Additional information

Used simplifier rules: fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m < # ar → ar[m, a][m] = a; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s)