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Proof step 9 from proof union-in-01 (previous step)

r -- sk ∈ st ∨ r -- sk ∈ st0, r ∈ 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: left 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]