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Proof step 4 from proof superset-elem-in-01 (previous step)

∀ a. a ∈ st → a ∈ st0 ∧ st[a] = st0[a], b ∈ st ⊦ b ∈ st0

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 1 terms: b

Additional information

Used terms: b