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

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 5:
a2 ∈ s, a1 ∈ s, a0 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a3 ∈ s ↔ a3 = b ∨ a3 = a, a2 ∈ s ↔ a2 = b, a1 = a
Proof step 15:
a1 ∈ s, a0 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a2 ∈ s, a3 ∈ s ↔ a3 = b ∨ a3 = a, a2 ∈ s ↔ a2 = b, a1 = a

Rule argument

fma: a2 ∈ s

Additional information

Used formulas: a2 ∈ s