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Proof step 6 from proof ocs-put (previous step)

a = a0, m ≤ # x0, ∀ m, a. m < # x0 → #oc(a, x0[m, a]) = #oc(a, x0) + 1 ∨ x0[m] = a ∨ x0 = [] ⊦ #oc(a, (a0 ' + x0)[m, a]) = (#oc(a, a0 ') + #oc(a, x0)) + 1, (a0 ' + x0)[m] = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
m ≤ # x0, ∀ m, a. m < # x0 → #oc(a, x0[m, a]) = #oc(a, x0) + 1 ∨ x0[m] = a ∨ x0 = [] ⊦ #oc(a0, (a0 ' + x0)[m, a0]) = 2 + #oc(a0, x0), (a0 ' + x0)[m] = a0

Rule argument

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic