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Proof step 9 from proof init-at-01 (previous step)

i ≤ n→i(# vals0) + j, j ≤ i, ∀ j0, i0, r0, st0. ¬ i0 < j0 ∧ ¬ j0 + n→i(# vals0) ≤ i0 → (addarray(r0, vals0, j0, st0)[r0 -- i0 ']) = vals0[i→n(i0 - j0)] ⊦ ((i = j ⊃ val; addarray(r, vals0, j + 1, st)[r -- i '])) = (val ' + vals0)[i→n(i - j)]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 8:
i = i1 + j, i ≤ n→i(# vals0) + j, j ≤ i, ∀ j0, i0, r0, st0. ¬ i0 < j0 ∧ ¬ j0 + n→i(# vals0) ≤ i0 → (addarray(r0, vals0, j0, st0)[r0 -- i0 ']) = vals0[i→n(i0 - j0)] ⊦ ((i = j ⊃ val; addarray(r, vals0, j + 1, st)[r -- i '])) = (val ' + vals0)[i→n(i1)]

Rule argument

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

Additional information

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