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Proof step 61 from proof addarray-addaray (previous step)

0 ≤ i ⊦ rk .ref = r, rk ∈ addarray(r, ty, i, st) ↔ rk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)), st)

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 59:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, i, st) ↔ r -- sk ∈ st)
Proof step 60:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, vals, st) ↔ r -- sk ∈ st)

Rule argument

[speclemmaarg: ,r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk ,[rk, r, sk] → [rk, r0, sk] ,T ,refkey , ,elim ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : ⊦ r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk])], [EXTRATERMS : (rk, r0, sk)]]