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Proof step 9 from proof ocs-same-length-mklist (previous step)

a = a0, ∀ a. #oc(a, x0) = # x0 ↔ x0 = mklist(a, # x0) ⊦ #oc(a, a0 ') + #oc(a, x0) = # x0 + 1 ↔ a0 = a ∧ x0 = mklist(a, # x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ a. #oc(a, x0) = # x0 ↔ x0 = mklist(a, # x0) ⊦ #oc(a0, x0) = # x0 ↔ x0 = mklist(a0, # x0)

Rule argument

Additional information

Used simplifier rules: one: ⊦ #oc(a, a ') = 1;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%eq-01: ⊦ m + n = m + n0 ↔ n = n0; from specification nat-basic2