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Proof step 4 from proof union-subset-04 (previous step)

a0 ∈ s, ∀ a. a ∈ s3 → a ∈ s4, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s → a ∈ s0 ⊦ s = ∅, a0 = a, # s = 0, a0 ∈ s4, a0 ∈ s2, a0 ∈ s0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a0

Additional information

Used terms: a0