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Proof step 63 from proof putarray-addarray-override (previous step)

0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0) ⊦ r0 -- sk ∈ putArray(vals, r, i, addarray(r, ty, vals0, st)) ↔ r0 -- sk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ rk ∈ putArray(vals, r, i, st) ↔ rk ∈ st ∨ rk .ref = r ∧ is_indexkey(rk .key) ∧ i ≤ rk .key .index ∧ rk .key .index < i + n→i(# vals)
Proof step 62:
0 ≤ i, i + n→i(# vals) ≤ n→i(# vals0) ⊦ r0 -- sk ∈ addarray(r, ty, vals0, st) ∨ (r0 -- sk) .ref = r ∧ is_indexkey((r0 -- sk) .key) ∧ i ≤ (r0 -- sk) .key .index ∧ (r0 -- sk) .key .index < i + n→i(# vals) ↔ r0 -- sk ∈ addarray(r, ty, firstn(i→n(i), vals0) + vals + restn(i→n(i) + # vals, vals0), st)

Rule argument

[speclemmaarg: , rk ∈ putArray(vals, r, i, st) ↔ rk ∈ st ∨ rk .ref = r ∧ is_indexkey(rk .key) ∧ i ≤ rk .key .index ∧ rk .key .index < i + n→i(# vals) ,[st, i, r, vals, rk] → [addarray(r, ty, vals0, st), i, r, vals, r0 -- sk] ,F ,storearrays , ,putArray-in ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addarray(r, ty, vals0, st), i, r, vals, r0 -- sk)]]