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

val = refval(r), validref(str, r, ty, st), is_referencevalue(val), ∀ val. val ∈ vals → is_referencevalue(val) ∧ validref(str, val .refval, ty, st), val0 = val ∨ val0 ∈ vals ⊦ is_referencevalue(val0) ∧ validref(str, val0 .refval, ty, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
validref(str, r, ty, st), ∀ val. val ∈ vals → is_referencevalue(val) ∧ validref(str, val .refval, ty, st), val0 = refval(r) ∨ val0 ∈ vals ⊦ is_referencevalue(val0) ∧ validref(str, val0 .refval, ty, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ is_referencevalue(refval(r))])]]