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Proof step 2 from proof in (previous step)

∀ a. a ∈ list2set(x0) ↔ a ∈ x0 ⊦ a0 ∈ list2set(a + x0) ↔ a0 ∈ a + x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ a. a ∈ list2set(x0) ↔ a ∈ x0 ⊦ a0 = a ∨ a0 ∈ list2set(x0) ↔ a0 = a ∨ a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: list2set-rec: ⊦ list2set(a ' + x) = list2set(x) ++ a;% ⊦ ¬ a ∈ b ' + x ↔ a ≠ b ∧ ¬ a ∈ x;% ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);% ⊦ (x + y) + z = x + y + z;% ⊦ a + x = a ' + x;