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Proof step 36 from proof putarray-addarray-one (previous step)

r0 -- sk ∈ addarray(r0, ty, n→i(# vals) + 1, st) ∨ ¬ sk .index ≤ 0 ∧ ¬ n→i(# vals) < sk .index ↔ r0 -- sk ∈ addarray(r0, ty, refval(jvmref) ' + vals, st), is_reftype(ty), is_indexkey(sk) ⊦ (putArray(vals, r0, 1, addarray(r0, ty, n→i(# vals) + 1, st))[r0 -- sk]) = (addarray(r0, ty, refval(jvmref) ' + vals, st)[r0 -- sk])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 35:
sk = i ', r0 -- sk ∈ addarray(r0, ty, n→i(# vals) + 1, st) ∨ ¬ i ≤ 0 ∧ ¬ n→i(# vals) < i ↔ r0 -- sk ∈ addarray(r0, ty, refval(jvmref) ' + vals, st), is_reftype(ty), is_indexkey(sk) ⊦ (putArray(vals, r0, 1, addarray(r0, ty, n→i(# vals) + 1, st))[r0 -- sk]) = (addarray(r0, ty, refval(jvmref) ' + vals, st)[r0 -- sk])

Rule argument

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

Additional information

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