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Proof step 2 from proof indef (previous step)

⊦ a ∈ s ↔ (∃ s0. (∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0) ∧ ¬ a ∈ s0)

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 3:
a ∈ s ⊦ a ∈ s ↔ (∃ s0. (∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0) ∧ ¬ a ∈ s0)
Proof step 8:
⊦ a ∈ s ↔ (∃ s0. (∀ a0. a0 ∈ s ↔ a0 = a ∨ a0 ∈ s0) ∧ ¬ a ∈ s0), a ∈ s

Rule argument

fma: a ∈ s

Additional information

Used formulas: a ∈ s