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Proof step 15 from proof hasSharing-rec (previous step)

hasSharing(r, st) ∨ (∃ r. r ∈ refs ∧ hasSharing(r, st)) ⊦ hasSharing(r ' + refs, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]