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Proof step 3 from proof del-two (previous step)

a3 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a2 ∈ s ↔ a2 = a, a1 ∈ s ↔ a1 = b, a0 ∈ s ↔ a0 ∈ ∅ ++ a ++ b

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
a4 ∈ s, a3 ∈ s, a1 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a0 ∈ s ↔ a0 ∈ ∅ ++ a ++ b, a1 ∈ s ↔ a1 = b, a2 ∈ s ↔ a2 = a
Proof step 16:
a4 ∈ s, a3 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a0 ∈ s ↔ a0 ∈ ∅ ++ a ++ b, a1 ∈ s ↔ a1 = b, a2 ∈ s ↔ a2 = a, a1 ∈ s

Rule argument

fma: a1 ∈ s

Additional information

Used formulas: a1 ∈ s