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Proof step 8 from proof newrefs-firstn (previous step)

n ≤ # refs, is_newref_list(refs, st) ⊦ ¬ dups(firstn(n, refs)) ∧ (∀ r. r ∈ firstn(n, refs) → newref(r, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ # refs, is_newref_list(refs, st) ⊦ ¬ dups(firstn(n, refs)) ∧ (r ∈ firstn(n, refs) → newref(r, st))

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]