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Proof step 13 from proof del-eq (previous step)

a ∈ s, a ∈ s0, a0 ∈ s, a1 ∈ s0, ∀ 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 14:
a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, a2 ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦ a2 ∈ s ↔ a2 ∈ s0, a2 = a
Proof step 16:
a1 ∈ s0, a0 ∈ s, a ∈ s0, a ∈ s, ∀ a. a ∈ s ↔ a ∈ s0 ⊦ a2 ∈ s ↔ a2 ∈ s0, a2 = a, a2 ∈ s

Rule argument

fma: a2 ∈ s

Additional information

Used formulas: a2 ∈ s