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Proof step 17 from proof del-two (
previous step
)
a0 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a1 ∈ s, a3 ∈ s ↔ a3 = b ∨ a3 = a, a2 ∈ s ↔ a2 = b, a1 ∈ s ↔ a1 = a
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 18
:
a4 ∈ s, a0 ∈ s, ∀ a1. a1 = b ∨ ¬ a1 ∈ s ⊦ a2 ∈ s ↔ a2 = b, a3 ∈ s ↔ a3 = b ∨ a3 = a, a ∈ s
Rule argument
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