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Proof step 7 from proof multi-subset (previous step)

# refs = i→n(countrefs(i ' + ints0)), 0 ≤ i0, posints(ints0), is_newref_list(refs, st), 0 ≤ i, r -- sk ∈ st, i ≤ i0, r ∈ st, r ∈ refs ⊦ st[r -- sk] = (addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st)[r -- sk]), 0 ∈ (i0 ' + ints0) .butlast

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ is_newref_list(refs, st) ↔ ¬ dups(refs) ∧ (∀ r. r ∈ refs → newref(r, st))
Proof step 6:
# refs = i→n(countrefs(i ' + ints0)), 0 ≤ i0, posints(ints0), ¬ dups(refs) ∧ (∀ r. r ∈ refs → newref(r, st)), 0 ≤ i, r -- sk ∈ st, i ≤ i0, r ∈ st, r ∈ refs ⊦ st[r -- sk] = (addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st)[r -- sk]), 0 ∈ (i0 ' + ints0) .butlast

Rule argument

[speclemmaarg: , is_newref_list(refs, st) ↔ ¬ dups(refs) ∧ (∀ r. r ∈ refs → newref(r, st)) ,[st, refs] → [st, refs] ,F ,storefuns , ,list_ok ]

Additional information

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