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Proof step 3 from proof in (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a0 ∈ x0, ∀ a. a ∈ list2set(x0) ↔ a ∈ x0 ⊦ a0 = a ∨ a0 ∈ list2set(x0) ↔ a0 = a ∨ a0 ∈ x0
Proof step 6:
∀ a. a ∈ list2set(x0) ↔ a ∈ x0 ⊦ a0 ∈ x0, a0 = a ∨ a0 ∈ list2set(x0) ↔ a0 = a ∨ a0 ∈ x0

Rule argument

fma: a0 ∈ x0

Additional information

Used formulas: a0 ∈ x0