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Proof step 2 from proof disjoint-in-left (previous step)

a ∈ st1, a ∈ st0, st0 ⊥ st1 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
a ∈ st0, a ∈ st1, st0 ⊥ st1 ⊦

Rule argument

Additional information