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Proof step 7 from proof del-two (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 8:
a4 ∈ s, a3 ∈ s, a1 ∈ s, a2 ∈ s, a0 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a0 = b, a0 = a, a1 = b, a2 = a
Proof step 9:
a4 ∈ s, a3 ∈ s, a1 ∈ s, a2 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s, a0 = b ∨ a0 = a ⊦ a1 = b, a2 = a, a0 ∈ s

Rule argument

pos: right 1

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