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Proof step 178 from proof addarray-put-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 177:
i→n(i) = n + i→n(j) ∧ i→n(j) ≤ i→n(i), 1 < i0 + j, 1 < i, 0 < i, j < i, i ≤ j + i0, 0 ≤ i, i ≤ i0 + j, 0 < j, 0 ≤ i0 + j, 0 ≤ i0, 0 < i0 ⊦ initial_value(ty) = (val ' + mklist(initial_value(ty), i→n(i0)))[n], j = 0, i = j

Rule argument

[speclemmaarg: n ≤ m ,n0 = m - n ↔ m = n0 + n ,[m, n, n0] → [i→n(i), i→n(j), n] ,T ,nat , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i→n(i) ≤ i→n(j) ↔ abs(i) ≤ abs(j)], [CSIMPSEQ : ⊦ 0 < j → abs(j) = j], [CSIMPSEQ : ⊦ 1 < i → abs(i) = i], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPSEQ : ⊦ i < j → ¬ j < i], [CSIMPELIM : n ≤ m ⊦ n0 = m - n ↔ m = n0 + n])], [EXTRATERMS : (i→n(i), i→n(j), n)]]