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Proof step 4 from proof in-append (previous step)

∀ a, y. a ∈ x0 + y ↔ a ∈ x0 ∨ a ∈ y ⊦ a ∈ x0 + y ↔ a ∈ x0 ∨ a ∈ y, a = a0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a ∈ x0, ∀ a, y. a ∈ x0 + y ↔ a ∈ x0 ∨ a ∈ y ⊦ a ∈ x0 + y ↔ a ∈ x0 ∨ a ∈ y, a = a0
Proof step 7:
∀ a, y. a ∈ x0 + y ↔ a ∈ x0 ∨ a ∈ y ⊦ a ∈ x0, a ∈ x0 + y ↔ a ∈ x0 ∨ a ∈ y, a = a0

Rule argument

fma: a ∈ x0

Additional information

Used formulas: a ∈ x0