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Proof step 19 from proof size-union (previous step)

a0 ∈ s0, a0 ∈ s1, ∀ a0. ¬ a0 ∈ s1 ∨ ¬ a0 ∈ s0 ⊦ s0 = ∅, s1 = ∅, #(s0 ∪ s1) = # s1 + # s0, # s0 = 0, # s1 = 0, a ∈ s1, a ∈ s0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a0

Additional information

Used terms: a0