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Proof step 17 from proof union-in-01 (previous step)

⊦ (∃ sk. r -- sk ∈ st ∪ st0) ↔ (∃ sk. r -- sk ∈ st) ∨ (∃ sk. r -- sk ∈ st0)

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]