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Proof step 2 from proof ocs-put-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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