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Proof step 13 from proof ocs-del-length (previous step)

∀ a, b. #oc(a, x0 -l b) = #oc(a, x0) ∨ a = b ⊦ #oc(a, a0 ' + (x0 -l b)) = #oc(a, x0), a = b, a = a0, b = a0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
∀ a, b. #oc(a, x0 -l b) = #oc(a, x0) ∨ a = b ⊦ #oc(a, x0 -l b) = #oc(a, x0), a = a0, a = b, b = a0

Rule argument

Additional information

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