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Proof step 17 from proof del-two (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 18:
a3 = a, a3 ∈ s, a4 ∈ s, ∀ a1. a1 = a ∨ ¬ a1 ∈ s ⊦ a0 ∈ s ↔ a0 ∈ ∅ ++ a ++ b, a2 ∈ s ↔ a2 = a, b ∈ s
Proof step 22:
a3 ∈ s, a4 ∈ s, ∀ a1. a1 = a ∨ ¬ a1 ∈ s ⊦ a0 ∈ s ↔ a0 ∈ ∅ ++ a ++ b, a2 ∈ s ↔ a2 = a, a3 = a, b ∈ s

Rule argument

fma: a3 = a

Additional information

Used formulas: a3 = a