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Proof step 15 from proof validreflist-rec (previous step)

⊦ (∀ val0. val0 ∈ val ' + vals → is_referencevalue(val0) ∧ validref(str, val0 .refval, ty, st)) ↔ is_referencevalue(val) ∧ validref(str, val .refval, ty, st) ∧ validreflist(str, vals, ty, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
⊦ (∀ 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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ ¬ val ∈ val0 ' + vals ↔ val ≠ val0 ∧ ¬ val ∈ vals])]]