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Proof step 7 from proof putArray-in (previous step)

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

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 6:
sk = i0 ', is_indexkey(sk), r -- sk ∈ putArray(vals0, r, i + 1, st[r -- i ', val]), r ∈ putArray(vals0, r, i + 1, st[r -- i ', val]), ∀ 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), i < i0 ∧ i0 ≤ i + n→i(# vals0) ∨ sk = i ' ⊦ r -- sk ∈ st, ¬ n→i(# vals0) + i < i0 ∧ ¬ i0 < i

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)]]