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Proof step 26 from proof del-two (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 27:
a3 ∈ s, a4 ∈ s, a0 ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦ a3 = a, a2 ∈ s ↔ a2 = b, a3 ∈ s ↔ a3 = b, a ∈ s
Proof step 29:
a4 ∈ s, a0 ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦ a3 ∈ s, a3 = a, a2 ∈ s ↔ a2 = b, a3 ∈ s ↔ a3 = b, a ∈ s

Rule argument

fma: a3 ∈ s

Additional information

Used formulas: a3 ∈ s