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Proof step 5 from proof in-1dim-02 (previous step)

0 < i, ∀ st, val, sk, r0, r, i0. i→n(i0) < i→n(i) → (true → r = r0 ∨ (r -- sk ∈ addarray(r0, val, i0, st) ↔ r -- sk ∈ st)) ⊦ r = r0, r -- sk ∈ addarray(r0, val, i - 1, st) ↔ r -- sk ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 4:
i = i1 + 1, 0 < i, ∀ st, val, sk, r0, r, i0. i→n(i0) < i→n(i) → (true → r = r0 ∨ (r -- sk ∈ addarray(r0, val, i0, st) ↔ r -- sk ∈ st)) ⊦ r = r0, r -- sk ∈ addarray(r0, val, i1, st) ↔ r -- sk ∈ 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)]]