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Proof step 14 from proof junkreflist-rec (previous step)

⊦ (∃ r. (refval(r) = val ∨ refval(r) ∈ vals) ∧ junkref(str, r, st)) ↔ is_referencevalue(val) ∧ junkref(str, val .refval, st) ∨ (∃ r. refval(r) ∈ vals ∧ junkref(str, r, st))

User applied rule case distinction

And obtained the following new goals

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]