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Proof step 4 from proof del-e-02 (previous step)

a1 ∈ s, a2 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ a0 ∈ s ↔ a0 = a

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a0 ∈ s, a1 ∈ s, a2 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ a0 ∈ s ↔ a0 = a
Proof step 7:
a1 ∈ s, a2 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ a0 ∈ s, a0 ∈ s ↔ a0 = a

Rule argument

fma: a0 ∈ s

Additional information

Used formulas: a0 ∈ s