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Proof step 4 from proof ocappend (previous step)

a0 = a, ∀ a, y. #oc(a, x0 + y) = #oc(a, x0) + #oc(a, y) ⊦ #oc(a, (a0 ' + x0) + y) = #oc(a, a0 ' + x0) + #oc(a, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ a, y. #oc(a, x0 + y) = #oc(a, x0) + #oc(a, y) ⊦ #oc(a, x0 + y) = #oc(a, x0) + #oc(a, y)

Rule argument

Additional information

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