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Proof step 13 from proof subseteq-rem-del (previous step)

a ∈ s0, a0 ∈ s0 → a0 ∈ s1, a ∈ s1, a0 = a ∨ a0 ∈ s1 ∧ ¬ a0 ∈ s0 ⊦ s0 = ∅, # s1 = 0, s1 = ∅, # s0 = 0, a0 ∈ s1 ∧ (a0 = a ∨ ¬ a0 ∈ s0)

User applied rule case distinction

And closed the goal!

Rule argument

pos: left 4

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