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Proof step 5 from proof del-two (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 6:
a4 ∈ s, a3 ∈ s, a1 ∈ s, a2 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a2 ∈ s ↔ a2 = a, a0 ∈ s ↔ a0 ∈ ∅ ++ a ++ b, a1 = b
Proof step 14:
a4 ∈ s, a3 ∈ s, a1 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a2 ∈ s ↔ a2 = a, a0 ∈ s ↔ a0 ∈ ∅ ++ a ++ b, a1 = b, a2 ∈ s

Rule argument

fma: a2 ∈ s

Additional information

Used formulas: a2 ∈ s