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Proof step 30 from proof union-addarray-rew (previous step)

0 ≤ sk .index, sk .index < n→i(# vals), is_indexkey(sk), newref(r0, st0), newref(r0, st) ⊦ r0 = jvmref, vals = [], (r0 -- sk ∈ addarray(r0, ty, vals, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addarray(r0, ty, vals, st0)) ∧ (addarray(r0, ty, vals, st ∪ st0)[r0 -- sk]) = (st ∪ addarray(r0, ty, vals, st0))[r0 -- sk]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 29:
sk = i ', 0 ≤ i, i < n→i(# vals), is_indexkey(sk), newref(r0, st0), newref(r0, st) ⊦ r0 = jvmref, vals = [], (r0 -- sk ∈ addarray(r0, ty, vals, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addarray(r0, ty, vals, st0)) ∧ (addarray(r0, ty, vals, st ∪ st0)[r0 -- sk]) = (st ∪ addarray(r0, ty, vals, st0))[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)]]