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Proof step 3 from proof notin (previous step)

∀ a. #oc(a, x0) = 0 ∨ a ∈ x0 ⊦ #oc(a, a0 + x0) = 0, a ∈ a0 + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ a. #oc(a, x0) = 0 ∨ a ∈ x0 ⊦ a = a0, #oc(a, x0) = 0, a ∈ x0

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;%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%in-06: ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x; from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list