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Proof step 4 from proof del-eq (previous step)

a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a2 ∈ s ↔ a2 ∈ s0

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a2 ∈ s, a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a2 ∈ s ↔ a2 ∈ s0
Proof step 7:
a ∈ s0, a ∈ s, a0 ∈ s0, a1 ∈ s, ∀ a0. a0 ≠ a → (a0 ∈ s ↔ a0 ∈ s0) ⊦ a2 ∈ s, a2 ∈ s ↔ a2 ∈ s0

Rule argument

fma: a2 ∈ s

Additional information

Used formulas: a2 ∈ s