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Proof step 5 from proof ocs-dups-ex (previous step)

∃ a, x0, y, z. x = x0 + (a + (y + (a + z))) ⊦ ∃ a. a ∈ x ∧ #oc(a, x) ≠ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ ∃ a0. (a0 ∈ x0 ∨ a0 = a ∨ a0 ∈ y ∨ a0 ∈ z) ∧ #oc(a0, x0) + #oc(a0, a ') + #oc(a0, y) + #oc(a0, a ') + #oc(a0, z) ≠ 1

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%in-01: ⊦ a ≠ b → (a ∈ b ' + x ↔ a ∈ x); from specification list%in-rec: ⊦ a ∈ b ' + x ↔ ¬ ¬(a = b ∨ a ∈ x); from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list