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Proof step 18 from proof del-two (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 19:
a3 = a, a4 ∈ s, a0 ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦ a2 ∈ s ↔ a2 = b, a3 ∈ s ↔ a3 ∈ ∅ ++ a ++ b, a ∈ s
Proof step 25:
a4 ∈ s, a0 ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦ a3 = a, a2 ∈ s ↔ a2 = b, a3 ∈ s ↔ a3 ∈ ∅ ++ a ++ b, a ∈ s

Rule argument

fma: a3 = a

Additional information

Used formulas: a3 = a