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Proof step 6 from proof validreflist-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
is_referencevalue(val), validref(str, val .refval, ty, st), val0 ∈ vals, ∀ val0. val0 ∈ vals → is_referencevalue(val0) ∧ validref(str, val0 .refval, ty, st) ⊦ is_referencevalue(val0) ∧ validref(str, val0 .refval, ty, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]