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Proof step 11 from proof ocs-same-length-mklist (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
#oc(a, x0) = 1 + # x0, ∀ a. #oc(a, x0) = # x0 ↔ x0 = mklist(a, # x0) ⊦ a = a0

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)