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Proof step 10 from proof ocs-sw (previous step)

¬(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]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
n < # x, m < # x ⊦ x[n] = x[m], #oc(x[m], x[m, x[n]][n, x[m]]) = #oc(x[m], x), x = []

Rule argument

Additional information

Used simplifier rules: put-same: ⊦ x[m] = a ∧ m < # x → x[m, a] = x; from specification list-del%put-length: ⊦ n < # x → # x[n, a] = # x; from specification list-del%put-at: m < # x ⊦ m ≠ n → x[m, a][n] = x[n]; from specification list-del%ls-le-simp: ⊦ m < n → (n ≤ m ↔ false); from specification nat%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat