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Proof step 11 from proof sorted-next-1-swap (previous step)

x[2 + 2 * n0] < x[2 * n0 + 1, x[2 + 2 * n0]][1 + 2 * n0], x[2 * n0 + 1] < x[2 * n0 + 1, x[2 + 2 * n0]][1 + 2 * n0], x[2 + 2 * n0] < x[2 * n0 + 1], 2 + 2 * n0 < # x, 2 * n0 + 1 < # x, 2 * n0 < # x, 2 + 2 * n0 < # x[2 * n0 + 1, x[2 + 2 * n0]], ∀ n. n < n0 → ¬ x[2 + 2 * n] < x[1 + 2 * n] ⊦ 2 + 2 * n0 = 1 + 2 * n0, x = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
2 * n0 < # x, 2 * n0 + 1 < # x, x[2 + 2 * n0] < x[2 * n0 + 1], x[2 * n0 + 1] < x[2 * n0 + 1, x[2 + 2 * n0]][1 + 2 * n0], 2 + 2 * n0 < # x, x[2 + 2 * n0] < x[2 * n0 + 1, x[2 + 2 * n0]][1 + 2 * n0], ∀ n. n < n0 → ¬ x[2 + 2 * n] < x[1 + 2 * n] ⊦ x = []

Rule argument

Additional information

Used simplifier rules: transitivity: ⊦ a < b ∧ b < c → a < c; (forward) from specification oelem% ⊦ n < # x → # x[n, a] = # x; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m + n = m + n0 ↔ n = n0; (s)