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Proof step 5 from proof del-e (previous step)

⊦ (∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ b) ↔ (∀ a0. ¬ a0 ∈ s), b ∈ s

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 6:
∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ b ⊦ b ∈ s, ∀ a0. ¬ a0 ∈ s
Proof step 8:
∀ a0. ¬ a0 ∈ s ⊦ b ∈ s, ∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ b

Rule argument

pos: right 1

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