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Proof step 10 from proof get-get (previous step)

n < i→n(i0), n→i(n) ≤ i0, 0 < i0, ∀ j0, n, r, i, st. i→n(j0) < i→n(i0 + 1) → (n→i(n) < j0 → (getarray(r, i, j0, st)[n]) = st[r -- (i + n→i(n)) ']) ⊦ n = i→n(i0), (getarray(r, i, i0, st)[n]) = st[r -- (i + n→i(n)) '], i0 = 0

User applied rule apply induction

And closed the goal!

Rule argument

[substlist-arg: [j0, n, r, i, st] → [i0, n, r, i, st]]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i→n(n→i(n)) = n], [CSIMPSEQ : ⊦ n→i(n) = 0 ↔ n = 0], [CSIMPFORWARD : ⊦ i < j ∧ j = i0 → i < i0], [CSIMPSEQ : ⊦ i ≤ j → (j ≤ i ↔ i = j)], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [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 : (i0, n, r, i, st)]]