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Proof step 4 from proof int-ins (previous step)

∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0 ++ a ⊦ (¬ a0 ∈ s ∨ ¬ a0 ∈ s0) ∧ ¬ a ∈ s

User applied rule case distinction

And obtained the following new goals

Proof step 5:
a0 ∈ s, a0 ∈ s0, ∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0 ++ a ⊦
Proof step 7:
a ∈ s, ∀ a0. ¬ a0 ∈ s ∨ ¬ a0 ∈ s0 ++ a ⊦

Rule argument

pos: right 1

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