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Proof step 5 from proof ocs-same-length-mklist (previous step)

1 + # x0 ≠ 0, ∀ a. #oc(a, x0) = # x0 ↔ x0 = mklist(a, # x0) ⊦ #oc(a, a0 ') + #oc(a, x0) = # x0 + 1 ↔ a0 ' + x0 = a + mklist(a, 1 + # x0 - 1)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
∀ a. #oc(a, x0) = # x0 ↔ x0 = mklist(a, # x0) ⊦ #oc(a, a0 ') + #oc(a, x0) = # x0 + 1 ↔ a0 ' + x0 = a ' + mklist(a, # x0)

Rule argument

Additional information

Used simplifier rules: oz: ⊦ m + 1 ≠ 0; from specification nat-basic2%c: ⊦ a + x = a ' + x; from specification list%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2%add-01: ⊦ n ≤ m → m + n0 - n = (m - n) + n0; from specification nat