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Proof step 11 from proof putArray-in (previous step)

is_indexkey(sk), ∀ rk, r, i, st. rk ∈ putArray(vals0, r, i, st) ↔ rk ∈ st ∨ rk .ref = r ∧ is_indexkey(rk .key) ∧ i ≤ rk .key .index ∧ rk .key .index < i + n→i(# vals0) ⊦ sk = sk .index ', r -- sk ∈ st, r -- sk ∈ putArray(vals0, r, sk .index + 1, st[r -- sk .index ', val])

Heuristic elimination applied rule insert elim lemma

And closed the goal!

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)]]