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Proof step 5 from proof newref-same (
previous step
)
r ∈ st ↔ 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 4
:
⊦ r ∈ st .refs ↔ r ∈ st
Rule argument
[crewrite-arg: refs-def, F, [st, r] → [st0, r]]
Additional information
[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0, r)]]