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Proof step 14 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) ∧ validreflist(str, vals, ty, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ validreflist(str, vals, ty, st) ↔ (∀ val. val ∈ vals → is_referencevalue(val) ∧ validref(str, val .refval, ty, st))
Proof step 13:
⊦ (∀ 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))

Rule argument

[crewrite-arg: validreflist-def , F , [st, ty, vals, str] → [st, ty, vals, str] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, ty, vals, str)]]