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Proof step 27 from proof addarray-addaray (previous step)

0 ≤ i, is_indexkey(sk), sk .index < i, 0 ≤ sk .index, 0 < i ⊦ (r -- sk ∈ addarray(r, ty, i, st) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ (addarray(r, ty, i, st)[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)[r -- sk])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 26:
sk = i0 ', 0 ≤ i, is_indexkey(sk), i0 < i, 0 ≤ i0, 0 < i ⊦ (r -- sk ∈ addarray(r, ty, i, st) ↔ r -- sk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)) ∧ (addarray(r, ty, i, st)[r -- sk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)[r -- sk])

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)]]