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Proof step 3 from proof init-at-01 (previous step)

(addarray(r, vals0, j + 1, st)[r -- (i1 + j) ']) = vals0[i→n(i1 - 1)], 0 < i1, i1 ≤ n→i(# vals0) ⊦ i1 = 0, vals0[i→n(i1 - 1)] = (val ' + vals0)[i→n(i1)], vals0 = []

Heuristic elimination applied rule insert elim lemma

And closed the goal!

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [i1, 1, i] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ((* j)) = i0 + ((* j0)) ↔ j - j0 = i0], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ 0 < i + 1 → i→n(i + 1) = i→n(i) + 1], [CSIMPSEQ : ⊦ (val ' + vals)[n + 1] = vals[n]], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (i1, 1, i)]]