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

∀ val0. val0 = val ∨ 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))

User applied rule all left

And obtained the following new goals

Proof step 6:
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))

Rule argument

[EXRARG : pos: left 1, [extquanttermlist: (val)abort:F, discard:F, computed:T]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val)]]