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Proof step 1 from proof addarrays-at-other-01 (
previous step
)
⊦ ¬ r ∈ refs ∧ # refs = i→n(countrefs(ints)) ∧ ints ≠ [] ∧ 0 ≤ i ∧ posints(ints) → (addarrays(refs, ty, ints, i, st)[r -- sk]) = st[r -- sk]
Lemma
addarrays-at-other