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Proof step 7 from proof junkreflist-rec (previous step)

val = refval(r0), junkref(str, r0, st), is_referencevalue(val) ⊦ ∃ r. (refval(r) = val ∨ refval(r) ∈ vals) ∧ junkref(str, r, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
junkref(str, r0, st) ⊦ ∃ r. (r = r0 ∨ refval(r) ∈ vals) ∧ junkref(str, r, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ refval(r) = refval(r0) ↔ r = r0], [CSIMPSEQ : ⊦ is_referencevalue(refval(r))])]]