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Proof step 14 from proof addarray-is-putarray (previous step)

0 ≤ i, 0 < i, ∀ st, val, r, i0. i→n(i0) < i→n(i) → (0 ≤ i0 → addarray(r, val, i0, st) = putArray(mklist(val, i→n(i0)), r, 0, st)) ⊦ (addarray(r, val, i - 1, st)[r -- (i - 1) ', val]) = putArray(mklist(val, i→n(i)), r, 0, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 13:
i = i1 + 1, 0 ≤ i, 0 < i, ∀ st, val, r, i0. i→n(i0) < i→n(i) → (0 ≤ i0 → addarray(r, val, i0, st) = putArray(mklist(val, i→n(i0)), r, 0, st)) ⊦ (addarray(r, val, i1, st)[r -- i1 ', val]) = putArray(mklist(val, i→n(i)), r, 0, st)

Rule argument

[speclemmaarg: ,i0 = i - j ↔ i = i0 + j ,[i, j, i0] → [i, 1, i1] ,T ,int-abs , ,elim-sub ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ i0 = i - j ↔ i = i0 + j])], [EXTRATERMS : (i, 1, i1)]]