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Proof step 2 from proof union-subset-01 (previous step)

a ∈ s, a0 ∈ s, ∀ a. a ∈ s → a ∈ s0 ⊦ a ∈ s1, a ∈ s0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

Additional information

Used terms: a