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Proof step 9 from proof instore-02 (
previous step
)
⊦ ∃ sk0. r0 -- sk0 ∈ st ∨ sk0 = sk
Heuristic quantifier closing applied rule exists right
And closed the goal!
Rule argument
[EXRARG : pos: right 1, [QUANTTERMLIST : (sk)]]
Additional information
[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk)]]