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Proof step 2 from proof ocs-put (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
m < # x ⊦ #oc(a, x[m, a]) = #oc(a, x) + 1, x[m] = a, x = []

Rule argument

Additional information

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