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Proof step 6 from proof refs-put (previous step)

r0 ∈ st[r -- sk, val] .refs ⊦ st[r -- sk, val] .refs = ∅, r0 = r, # st[r -- sk, val] .refs = 0, r0 ∈ st .refs

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ r ∈ st .refs ↔ r ∈ st
Proof step 5:
r0 ∈ st[r -- sk, val] ⊦ st[r -- sk, val] .refs = ∅, r0 = r, # st[r -- sk, val] .refs = 0, r0 ∈ st .refs

Rule argument

[crewrite-arg: refs-def, F, [st, r] → [st[r -- sk, val], r0]]

Additional information

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