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Proof step 6 from proof union-subset-03 (previous step)

a ∈ s1, a0 ∈ s1, ∀ a. a ∈ s3 → a ∈ s4, ∀ a. a ∈ s1 → a ∈ s2, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s4, a ∈ s2, a ∈ s0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 4 terms: a

Additional information

Used terms: a