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Proof step 10 from proof getarray-put-01 (previous step)

0 < j, i0 < i ∨ i + j ≤ i0, ∀ j0, i0, i, r, st, val. i→n(j0) < i→n(j) → (i0 < i ∨ i + j0 ≤ i0 → getarray(r, i, j0, st[r -- i0 ', val]) = getarray(r, i, j0, st)) ⊦ ((i = i0 ⊃ val; st[r -- i '])) ' + getarray(r, i + 1, j - 1, st[r -- i0 ', val]) = getarray(r, i, j, st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 9:
j = i1 + 1, 0 < j, i0 < i ∨ i + j ≤ i0, ∀ j0, i0, i, r, st, val. i→n(j0) < i→n(j) → (i0 < i ∨ i + j0 ≤ i0 → getarray(r, i, j0, st[r -- i0 ', val]) = getarray(r, i, j0, st)) ⊦ ((i = i0 ⊃ val; st[r -- i '])) ' + getarray(r, i + 1, i1, st[r -- i0 ', val]) = getarray(r, i, j, st)

Rule argument

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

Additional information

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