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Proof step 9 from proof ocs-sw (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 10:
⊦ x[m] = a ∧ a ≠ b ∧ m < # x → #oc(a, x[m, b]) = #oc(a, x) -1
Proof step 11:
n < # x, m < # x ⊦ #oc(a, x[m, x[n]][n, x[m]]) = #oc(a, x), m = n, x = [], x[n] ≠ a ∧ a ≠ x[m], x[m, x[n]][n] = a ∧ a ≠ x[m] ∧ n < # x[m, x[n]]

Rule argument

spec: <toplevel> 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: <all paths>

Additional information

Used simplifier rules: ocs-put: ⊦ x[m] ≠ a ∧ m < # x → #oc(a, x[m, a]) = #oc(a, x) + 1;% ⊦ m < n ↔ ¬ n ≤ m; (ss)% ⊦ m < n → (n ≤ m ↔ false); (s)%m < # x ⊦ m ≠ n → x[m, a][n] = x[n]; (s)% ⊦ n -1 = n - 1; (s)% ⊦ n ≤ m → m + n0 - n = (m - n) + n0; (s)% ⊦ n + 0 = n; (s)%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic%c: ⊦ m + n = n + m; from specification nat-basic Used terms: x[m], n, x[m, x[n]], a