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Proof step 90 from proof putarray-addarray-01 (previous step)

0 ≤ i, 0 ≤ i0 + n→i(# vals), 0 ≤ i0 + i + n→i(# vals), 0 < i0, is_indexkey(sk) ⊦ sk = _length, ( r -- sk ∈ addarray(r, ty, i0 + i + n→i(# vals), st) ∨ i ≤ sk .index ∧ sk .index < i + n→i(# vals) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)) ∧ (putArray(vals, r, i, addarray(r, ty, i0 + i + n→i(# vals), st))[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)[r -- sk])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 89:
sk = i1 ', 0 ≤ i, 0 ≤ i0 + n→i(# vals), 0 ≤ i0 + i + n→i(# vals), 0 < i0, is_indexkey(sk) ⊦ sk = _length, ( r -- sk ∈ addarray(r, ty, i0 + i + n→i(# vals), st) ∨ i ≤ i1 ∧ i1 < i + n→i(# vals) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)) ∧ (putArray(vals, r, i, addarray(r, ty, i0 + i + n→i(# vals), st))[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)[r -- sk])

Rule argument

[speclemmaarg: ,is_indexkey(sk) → (i = sk .index ↔ sk = i ') ,[sk, i] → [sk, i1] ,T ,storekey , ,elim-04 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ is_indexkey(sk) → (i = sk .index ↔ sk = i ')])], [EXTRATERMS : (sk, i1)]]