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Proof step 16 from proof junkreflist-rec (previous step)

⊦ (∃ r. refval(r) ∈ val ' + vals ∧ junkref(str, r, st)) ↔ is_referencevalue(val) ∧ junkref(str, val .refval, st) ∨ junkrefs(str, vals, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
⊦ (∃ r. (refval(r) = val ∨ refval(r) ∈ vals) ∧ junkref(str, r, st)) ↔ is_referencevalue(val) ∧ junkref(str, val .refval, st) ∨ junkrefs(str, vals, st)

Rule argument

[emptyarg]

Additional information

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