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Proof step 9 from proof getarray-addarray-03 (previous step)

0 ≤ i1, 0 < i1, ∀ st, val, r, i0. i→n(i0) < i→n(i1 + 1) → (0 ≤ i0 → getarray(r, 0, i0, addarray(r, val, i0, st)) = mklist(val, i→n(i0))) ⊦ getarray(r, 0, i1, addarray(r, val, i1, st)) = mklist(val, i→n(i1))

User applied rule apply induction

And closed the goal!

Rule argument

[substlist-arg: [st, val, r, i0] → [st, val, r, i1]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ 0 < i → i→n(i + 1) = i→n(i) + 1], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ m < n + 1 ↔ ¬ n < m], [CSIMPSEQ : ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j)], [CSIMPSEQ : ⊦ 0 < j → abs(j) = j], [CSIMPSEQ : ⊦ ¬ i < i], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRATERMS : (st, val, r, i1)]]