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

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ junkrefs(str, vals, st) ↔ (∃ r. refval(r) ∈ vals ∧ junkref(str, r, st))
Proof step 14:
⊦ (∃ r. (refval(r) = val ∨ refval(r) ∈ vals) ∧ junkref(str, r, st)) ↔ is_referencevalue(val) ∧ junkref(str, val .refval, st) ∨ (∃ r. refval(r) ∈ vals ∧ junkref(str, r, st))

Rule argument

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

Additional information

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