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Proof step 5 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))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
junkref(str, r, st), refval(r) = val ∨ refval(r) ∈ vals ⊦ ∃ r. refval(r) ∈ vals ∧ junkref(str, r, st), is_referencevalue(val) ∧ junkref(str, val .refval, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]