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Proof step 10 from proof hasSharing-rec (previous step)

hasSharing(r, st) ⊦ (r0 = r ∨ r0 ∈ refs) ∧ hasSharing(r0, st), ∃ r1. (r1 = r ∨ r1 ∈ refs) ∧ hasSharing(r1, st)

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

[EXRARG : pos: right 2, [QUANTTERMLIST : (r)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r)]]