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Proof step 2 from proof subsub-del (previous step)

a0 ∈ s, a ∈ s, a ∈ s0, a1 ∈ s, a2 ∈ s0, ∀ a0. a0 ∈ s → a0 ∈ s0 ⊦ a0 ∈ s0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: a0

Additional information

Used terms: a0