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Proof step 1 from proof ocs-put (conclusion)

⊦ x[m] ≠ a ∧ m < # x → #oc(a, x[m, a]) = #oc(a, x) + 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
m < # x ⊦ x[m] = a, #oc(a, x[m, a]) = #oc(a, x) + 1, x = []

Rule argument

Additional information

Used simplifier rules: l: ⊦ # x = 0 ↔ x = []; from specification list-data%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1