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Proof step 9 from proof shift-1dim-01 (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
i = i1 + 1, 0 < i, ∀ val, sk, st, val0, r0, r, i0. i→n(i0) < i→n(i) → (true → r = r0 ∨ (addarray(r, val0, i0, st)[r0 -- sk, val]) = addarray(r, val0, i0, st[r0 -- sk, val])) ⊦ (addarray(r, val0, i1, st)[r -- i1 ', val0][r0 -- sk, val]) = addarray(r, val0, i, st[r0 -- sk, val]), r = r0

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)]]