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Proof step 8 from proof int-ins (previous step)

a ∈ s, a0 ∈ s, ∀ a0. ¬ a0 ∈ s ∨ a0 ≠ a ∧ ¬ a0 ∈ s0 ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

Additional information

Used terms: a