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Proof step 14 from proof union-size-le (previous step)

# s0 + # s < #(s ∪ s0), a ∈ s, ∀ s0. # s0 + # s < #(s0 ∪ s) → #(s0 ∪ s) = 0 ∨ s0 ∪ s = ∅ ⊦ s = ∅, s ∪ s0 = ∅, #(s ∪ s0) = 0, # s = 0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: s0

Additional information

Used simplifier rules: unc: ⊦ s ∪ s0 = s0 ∪ s; Used terms: s0