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Proof step 20 from proof size-exgr (previous step)

∀ s. # s ≠ 0 → (∃ a0. a0 ∈ s) ⊦ s = ∅, # s = 0, a ∈ s, ∃ a0. a0 ∈ s

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: s

Additional information

Used terms: s