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Proof step 9 from proof del-two (previous step)

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

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 10:
a0 = b, a2 ∈ s, a1 ∈ s, a3 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a1 = b, a2 = a, a0 ∈ s
Proof step 12:
a0 = a, a2 ∈ s, a1 ∈ s, a3 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a1 = b, a2 = a, a0 ∈ s

Rule argument

pos: left 6

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