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Proof step 174 from proof addarray-put-01 (previous step)

i→n(i) = n + i→n(j), 1 < i0 + j, 1 < i, 0 < i, i ≤ j + i0, 0 ≤ i, i ≤ i0 + j, 0 < j, 0 ≤ i0 + j, 0 ≤ i0, 0 < i0, j < i ⊦ j = 0, i = j, initial_value(ty) = (mklist(initial_value(ty), i→n(i0))[n - 1]), n = 0

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 173:
n = m + 1, i→n(i) = n + i→n(j), 1 < i0 + j, 1 < i, 0 < i, i ≤ j + i0, 0 ≤ i, i ≤ i0 + j, 0 < j, 0 ≤ i0 + j, 0 ≤ i0, 0 < i0, j < i ⊦ j = 0, i = j, initial_value(ty) = (mklist(initial_value(ty), i→n(i0))[m]), n = 0

Rule argument

[speclemmaarg: n ≠ 0 ,m = n - 1 ↔ n = m + 1 ,[n, m] → [n, m] ,T ,nat , ,elim-pred-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (n, m)]]