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Proof step 20 from proof reffield-del (previous step)

has_reffield(r0, st -- r), is_valid_ref(r0, st) ⊦ r0 = r, has_reffield(r0, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ is_valid_ref(r, st) ↔ is_obj(r, st) ∨ is_array(r, st)
Proof step 19:
has_reffield(r0, st -- r), is_obj(r0, st) ∨ is_array(r0, st) ⊦ r0 = r, has_reffield(r0, st)

Rule argument

[crewrite-arg: is-valid-ref, F, [st, r] → [st, r0]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r0)]]