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Proof step 110 from proof putarray-addarray-01 (previous step)

0 ≤ i, 0 ≤ i0 + n→i(# vals), 0 ≤ i0 + i + n→i(# vals), 0 < i0 ⊦ rk = r -- _length, ( rk ∈ addarray(r, ty, i0 + i + n→i(# vals), st) ∨ rk .ref = r ∧ is_indexkey(rk .key) ∧ i ≤ rk .key .index ∧ rk .key .index < i + n→i(# vals) ↔ rk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)) ∧ (putArray(vals, r, i, addarray(r, ty, i0 + i + n→i(# vals), st))[rk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)[rk])

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 109:
rk = r0 -- sk, 0 ≤ i, 0 ≤ i0 + n→i(# vals), 0 ≤ i0 + i + n→i(# vals), 0 < i0 ⊦ rk = r -- _length, ( rk ∈ addarray(r, ty, i0 + i + n→i(# vals), st) ∨ r0 = r ∧ is_indexkey(sk) ∧ i ≤ sk .index ∧ sk .index < i + n→i(# vals) ↔ rk ∈ addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)) ∧ (putArray(vals, r, i, addarray(r, ty, i0 + i + n→i(# vals), st))[rk]) = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals + mklist(initial_value(ty), i→n(i0)), st)[rk])

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)]]