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Proof step 12 from proof validreflist-rec (previous step)

is_referencevalue(val) ∧ validref(str, val .refval, ty, st) ∧ (∀ val. val ∈ vals → is_referencevalue(val) ∧ validref(str, val .refval, ty, st)) ⊦ ∀ val0. 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 11:
validref(str, val .refval, 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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]