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Proof step 9 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 ⊦ 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 3:
⊦ ¬ r ∈ refs ∧ # refs = i→n(countrefs(i ' + ints .rest)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) ∧ ¬ 0 ∈ ints .butlast ∧ i ≤ ints .first → (addmultiarrays(refs, i, acc, ty, ints, st)[r -- sk]) = st[r -- sk]
Proof step 8:
¬ ( ¬ r ∈ refs ∧ # refs = i→n(countrefs(i ' + (i0 ' + ints0) .rest)) ∧ i0 ' + ints0 ≠ [] ∧ 0 ≤ i ∧ posints(i0 ' + ints0) ∧ ¬ 0 ∈ (i0 ' + ints0) .butlast ∧ i ≤ (i0 ' + ints0) .first), # refs = i→n(countrefs(i ' + ints0)), 0 ≤ i0, posints(ints0), is_newref_list(refs, st), 0 ≤ i, r -- sk ∈ st, i ≤ i0, r ∈ st ⊦ st[r -- sk] = (addmultiarrays(refs, i, acc, ty, i0 ' + ints0, st)[r -- sk]), 0 ∈ (i0 ' + ints0) .butlast

Rule argument

[crewrite-arg: multi-at-other , F , [sk, r, st, ints, ty, acc, i, refs] → [sk, r, st, i0 ' + ints0, ty, acc, i, refs] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, r, st, i0 ' + ints0, ty, acc, i, refs)]]