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Proof step 14 from proof ocs-sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
x[m, x[n]][n] = x[m], m < # x, n < # x ⊦ #oc(x[m], x[m, x[n]][n, x[m]]) = #oc(x[m], x), x[n] = x[m], x = []

Rule argument

Additional information

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