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Proof step 2 from proof in-05 (previous step)

∀ a0. a0 = a ∨ a0 ∈ s ↔ a0 = b ∨ a0 ∈ s0 ⊦ a = b, a ∈ s0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: a

Additional information

Used terms: a