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Proof step 8 from proof ocs-sw (previous step)

n < # x, m < # x ⊦ #oc(a, x[m, x[n]][n, x[m]]) = #oc(a, x), x = [], m = n, x[n] ≠ a ∧ a ≠ x[m]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ x[m] = a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x) -1
Proof step 10:
¬(x[m, x[n]][n] = a ∧ a ≠ x[m] ∧ n < # x[m, x[n]]), n < # x, m < # x ⊦ #oc(a, x[m, x[n]][n, x[m]]) = #oc(a, x), x = [], m = n, x[n] ≠ a ∧ a ≠ x[m]

Rule argument

spec: list-perm inst: name: ocs-put-02 seq: ⊦ x[m] = a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x) -1 subst: [b, m, x, a] → [x[m], n, x[m, x[n]], a] flags: rewrite, current

Additional information

Used simplifier rules: ocs-put: ⊦ x[m] ≠ a ∧ m < # x → #oc(a, x[m, a]) = #oc(a, x) + 1;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%put-at: m < # x ⊦ m ≠ n → x[m, a][n] = x[n]; from specification list-del%pred-03: ⊦ n -1 = n - 1; from specification nat%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat%add-base: ⊦ n + 0 = n; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: x[m], n, x[m, x[n]], a