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Proof step 8 from proof sorted-insert-merge (previous step)

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] ⊦ n0 = m0, A[2 + m0] < a, a < A[m0], A[2 + m0] < A[m0]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ m ≠ n → ar[n, a][m] = ar[m]
Proof step 10:
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

Rule argument

spec: array name: Put-other seq: ⊦ m ≠ n → ar[n, a][m] = ar[m] subst: [m, a, n, ar] → [n0, a, m0 + 1, A] flags: <all paths>

Additional information

Used simplifier rules: ⊦ m = m + n ↔ n = 0; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m + 1 ↔ n = m + 1); (s)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat% ⊦ m ≤ m; (s)% ⊦ m + n < m + n0 ↔ n < n0; (s)%m < # ar ⊦ ar[m, a][m] = a; (s) Used terms: n0, a, m0 + 1, A