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Proof step 13 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) ∧ (∀ val. val ∈ vals → is_referencevalue(val) ∧ validref(str, val .refval, ty, st))

User applied rule case distinction

And obtained the following new goals

Proof step 8:
∀ val0. val0 = val ∨ val0 ∈ vals → is_referencevalue(val0) ∧ validref(str, val0 .refval, ty, st) ⊦ is_referencevalue(val) ∧ validref(str, val .refval, ty, st) ∧ (∀ val. val ∈ vals → is_referencevalue(val) ∧ validref(str, val .refval, ty, st))
Proof step 12:
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)

Rule argument

[CASEDARG : pos: right 1, ()]

Additional information

[[CONCRETESIMPRULES : ()]]