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Proof step 23 from proof ocs-sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 24:
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-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ m = m + n ↔ n = 0; (s)