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Proof step 22 from proof ocs-sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 23:
n < # x, m < # x ⊦ x[m, x[n]][n] = x[n], x[n] = x[m], x = []

Rule argument

Additional information

Used simplifier rules: ocs-put: ⊦ x[m] ≠ a ∧ m < # x → #oc(a, x[m, a]) = #oc(a, x) + 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq: ⊦ m = m + n ↔ n = 0; from specification nat-basic2