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Proof step 6 from proof in (previous step)

∀ a. a ∈ list2set(x0) ↔ a ∈ x0 ⊦ a0 ∈ x0, a0 = a ∨ a0 ∈ list2set(x0) ↔ a0 = a ∨ a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a0 ∈ list2set(x0), ∀ a. a ∈ list2set(x0) ↔ a ∈ x0 ⊦ list2set(x0) = ∅, a0 = a, # list2set(x0) = 0, a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: ⊦ s ≠ ∅ → # s ≠ 0; (forward)% ⊦ a ∈ s → s ≠ ∅; (forward)