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Proof step 6 from proof del-two (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 7:
a3 = a, a2 ∈ s, a1 ∈ s, a0 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a1 = a, a3 ∈ s ↔ a3 ∈ ∅ ++ a ++ b, a2 = b
Proof step 9:
a2 ∈ s, a1 ∈ s, a0 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a3 = a, a1 = a, a3 ∈ s ↔ a3 ∈ ∅ ++ a ++ b, a2 = b

Rule argument

fma: a3 = a

Additional information

Used formulas: a3 = a