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Proof step 9 from proof newrefs-firstn (previous step)

n ≤ # refs, is_newref_list(refs, st) ⊦ is_newref_list(firstn(n, refs), st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ is_newref_list(refs, st) ↔ ¬ dups(refs) ∧ (∀ r. r ∈ refs → newref(r, st))
Proof step 8:
n ≤ # refs, is_newref_list(refs, st) ⊦ ¬ dups(firstn(n, refs)) ∧ (∀ r. r ∈ firstn(n, refs) → newref(r, st))

Rule argument

[crewrite-arg: list_ok, F, [st, refs] → [st, firstn(n, refs)]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, firstn(n, refs))]]