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Proof step 5 from proof sorted-next-1 (previous step)

n % 2 = 1, x[2 + 2 * n0] < x[1 + 2 * n0], 2 * n0 < n, n ≤ 2 + 2 * n0 ⊦ x[n + 1] < x[n]

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: nat name: anti seq: ⊦ m ≤ n → (n ≤ m ↔ m = n) subst: [m, n] → [2 + 2 * n0, n] flags: <all paths>

Additional information

Used simplifier rules: n ≠ 0 ⊦ (m + n0 * n) % n = m % n; (s)%mult-a: ⊦ m0 * m * n = (m0 * m) * n; from specification nat-mult%mult-c: ⊦ m * n = n * m; from specification nat-mult%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic% ⊦ m ≤ n → (n ≤ m ↔ m = n); (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)% ⊦ m < n → (n < m + 2 ↔ m + 1 = n); (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% ⊦ (2 * m + 1) % 2 = 1; (s)% ⊦ m + n = m + n0 ↔ n = n0; (s)% ⊦ m + n ≤ m + n0 ↔ n ≤ n0; (s)% ⊦ m < n + 1 ↔ ¬ n < m; (s)% ⊦ 2 * m < 2 * n ↔ m < n; (s)% ⊦ ¬ n < n; (s)%nll: ⊦ ¬ b < a ∧ b < c → a < c; (forward) from specification oelem Used terms: 2 + 2 * n0, n