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Proof step 22 from proof reffield-del (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ ¬ is_valid_ref(r, st) → ¬ has_reffield(r, st)
Proof step 21:
¬ ¬ is_valid_ref(r0, st -- r), has_reffield(r0, st -- r) ⊦ has_reffield(r0, st), r0 = r

Rule argument

[crewrite-arg: has_reffield-no, F, [st, r] → [st -- r, r0]]

Additional information

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