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Proof step 30 from proof putarray-addarray-override (previous step)

is_indexkey(sk), 0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0) ⊦ r0 -- sk ∈ addarray(r, ty, vals0, st) ∨ r0 = r ∧ i ≤ sk .index ∧ sk .index < i + n→i(# vals) ↔ r0 -- sk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 29:
sk = i0 ', is_indexkey(sk), 0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0) ⊦ r0 -- sk ∈ addarray(r, ty, vals0, st) ∨ r0 = r ∧ i ≤ i0 ∧ i0 < i + n→i(# vals) ↔ r0 -- sk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)

Rule argument

[speclemmaarg: ,is_indexkey(sk) → (i = sk .index ↔ sk = i ') ,[sk, i] → [sk, i0] ,T ,storekey , ,elim-04 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_indexkey(sk) → (i = sk .index ↔ sk = i ')])], [EXTRATERMS : (sk, i0)]]