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Proof step 9 from proof newref-same (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
st .refs = st0 .refs ⊦ r ∈ st ↔ r ∈ st0, newref(st) = newref(st0)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]