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Proof step 10 from proof del-eq (previous step)

a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦ a2 ∈ s ↔ a2 ∈ s0, a2 = a

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 11:
a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦ a2 = a, a2 ∈ s ↔ a2 ∈ s0, a2 = a

Rule argument

fma: a2 = a

Additional information

Used formulas: a2 = a