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Proof step 2 from proof addarray-mklist-04 (previous step)

0 ≤ i ⊦ addarray(r, mklist(val, n), i, addarray(r, mklist(val, i→n(i)), 0, st)) = addarray(r, val, n→i(n) + i, st)

User applied rule insert lemma

And obtained the following new goals

Proof step 1:
⊦ addarray(r, mklist(val, n), n→i(m), addarray(r, mklist(val, m), 0, st)) = addarray(r, val, n→i(n) + n→i(m), st)

Rule argument

[LEMMAARG : addarray-mklist-03, [r, val, n, m, st] → [r, val, n, i→n(i), st]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ 0 ≤ i → n→i(i→n(i)) = i])], [EXTRATERMS : (r, val, n, i→n(i), st)]]