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Proof step 5 from proof shift-1dim-01 (previous step)

0 ≤ i1, ∀ val, sk, st, val0, r0, r, i0. i→n(i0) < i→n(i1 + 1) → (true → r = r0 ∨ (addarray(r, val0, i0, st)[r0 -- sk, val]) = addarray(r, val0, i0, st[r0 -- sk, val])) ⊦ r = r0, (addarray(r, val0, i1, st)[r0 -- sk, val][r -- i1 ', val0]) = (addarray(r, val0, i1, st[r0 -- sk, val])[r -- i1 ', val0])

User applied rule apply induction

And closed the goal!

Rule argument

[substlist-arg: [val, sk, st, val0, r0, r, i0] → [val, sk, st, val0, r0, 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 ≤ i → abs(i) = i], [CSIMPSEQ : ⊦ ¬ i < i], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ m + n = n + m])], [EXTRATERMS : (val, sk, st, val0, r0, r, i1)]]