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Proof step 3 from proof in-05 (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:
#oc(a, x0) = 0, a ∈ x0, ∀ a1. #oc(a1, x0) ≠ 0 ∨ ¬ a1 ∈ x0 ⊦ a = a0

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%one: ⊦ #oc(a, a ') = 1;%one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ 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%z-01: ⊦ m + n = 0 ↔ m = 0 ∧ n = 0; from specification nat-basic2%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list