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Proof step 20 from proof violations-swap (previous step)

less#(a, x0) + violations(x0) = m + 1, m1 + 1 < # x0, x0[m1 + 1] < x0[m1], ∀ m0, n0. violations(x0) = m0 + 1 ∧ x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0 → violations(x0[n0, x0[n0 + 1]][n0 + 1, x0[n0]]) = m0 ⊦ less#(a, x0) + violations(x0[m1, x0[m1 + 1]][1 + m1, x0[m1]]) = m, x0 = []

User applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: m - less#(a, x0), m1 abort: F discard: T computed: F

Additional information

Used simplifier rules: violations-exists: n + 1 < # x ⊦ x[n + 1] < x[n] → violations(x) ≠ 0;% ⊦ m ≤ n ↔ ¬ n < m; (ss)%eqle-one: ⊦ m = n + 1 ∧ n < n0 → m ≤ n0; (forward) from specification nat%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)% ⊦ m + m0 ≤ m ↔ m0 = 0; (s)%n ≤ m ⊦ (m - n) + n0 = n1 ↔ m + n0 = n + n1; (s) Used terms: m - less#(a, x0), m1