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Proof step 16 from proof addarrays-subset (previous step)

r -- sk ∈ st, is_newref_list(refs, st), 0 ≤ i, r ∈ st ⊦ st[r -- sk] = (addarrays(refs, ty, ints, i, st)[r -- sk])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ r ∈ refs ∧ # refs = i→n(countrefs(ints)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) → (addarrays(refs, ty, ints, i, st)[r -- sk]) = st[r -- sk]
Proof step 15:
¬ (¬ r ∈ refs ∧ # refs = i→n(countrefs(ints)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints)), r -- sk ∈ st, is_newref_list(refs, st), 0 ≤ i, r ∈ st ⊦ st[r -- sk] = (addarrays(refs, ty, ints, i, st)[r -- sk])

Rule argument

[crewrite-arg: addarrays-at-other , F , [sk, r, st, i, ints, ty, refs] → [sk, r, st, i, ints, ty, refs] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, r, st, i, ints, ty, refs)]]