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Proof step 10 from proof newrefs-firstn (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
n ≤ # refs, is_newref_list(refs, st) ⊦ is_newref_list(firstn(n, refs), st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]