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

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

User applied rule case distinction

And obtained the following new goals

Proof step 5:
∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0 ⊦ ¬ a0 ∈ s ∨ ¬ a0 ∈ s0
Proof step 7:
∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0 ⊦ ¬ a ∈ s

Rule argument

pos: right 1

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