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Proof step 5 from proof ocs-sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
m < # x, n < # x ⊦ #oc(a, x[m, x[n]][n, x[m]]) = #oc(a, x), x = [], x[m, x[n]][n] ≠ a ∧ a ≠ x[m]

Rule argument

Additional information

Used simplifier rules: ⊦ n < # x → # x[n, a] = # x; (s)