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Proof step 6 from proof newref-same (previous step)

r ∈ st .refs ↔ r ∈ st0 .refs ⊦ r ∈ st ↔ r ∈ st0, newref(st) = newref(st0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ∈ st .refs ↔ r ∈ st
Proof step 5:
r ∈ st ↔ r ∈ st0 .refs ⊦ r ∈ st ↔ r ∈ st0, newref(st) = newref(st0)

Rule argument

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

Additional information

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