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Proof step 7 from proof disj (previous step)

r ∈ rs1, r ∈ rs0, ∀ r. ¬ r ∈ rs0 ∨ ¬ r ∈ rs1 ⊦ #(rs0 ∩ rs1) = 0, # rs1 = 0, # rs0 = 0, rs0 = ∅, rs1 = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: r

Additional information

Used terms: r