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Proof step 25 from proof subset-is-union (previous step)

s ⊆ s1, a ∈ s1, ∀ s. s ⊆ s1 → (∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅) ⊦ s1 = ∅, # s1 = 0, ∃ s2. s ∪ s2 = s1 ∧ s ∩ s2 = ∅

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: s

Additional information

Used terms: s