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Proof step 220 from proof putarray-addarray-01 (previous step)

j ≤ n→i(# vals) + i, i ≤ j, 0 ≤ i, 0 ≤ j, 0 ≤ n→i(# vals) + i ⊦ rk .ref = r, rk = r -- _length, (rk ∈ putArray(vals, r, i, addarray(r, ty, j, st)) ↔ rk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st)) ∧ (putArray(vals, r, i, addarray(r, ty, j, st))[rk]) = st[rk]

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 217:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, i, st) ↔ r -- sk ∈ st)
Proof step 218:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, vals, st) ↔ r -- sk ∈ st)
Proof step 219:
⊦ r ≠ r0 → (addarray(r, ty, i, st)[r0 -- sk]) = st[r0 -- sk]

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)], [CSIMPSEQ : ⊦ r ≠ r0 → ( r0 -- sk ∈ putArray(vals, r, j, st) ↔ r0 -- sk ∈ st)], [CSIMPSEQ : ⊦ r ≠ r0 → (putArray(vals, r, j, st) [r0 -- sk ] ) = st[r0 -- sk]], [CSIMPELIM : ⊦ r = rk .ref ∧ sk = rk .key ↔ rk = r -- sk])], [EXTRATERMS : (rk, r0, sk)]]