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Proof step 12 from proof subseteq-inter-empty-empty (previous step)

a ∈ s1, a ∈ s0, a ∈ s2, ∀ a. ¬ a ∈ s1 ∨ ¬ a ∈ s2, ∀ a. a ∈ s0 → a ∈ s1 ⊦ s1 = ∅, # s0 = 0, # s2 = 0, #(s0 ∩ s2) = 0, s0 = ∅, s2 = ∅, # s1 = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a

Additional information

Used terms: a