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Proof step 211 from proof putarray-addarray-01 (previous step)

is_fskey(sk), 0 ≤ n→i(# vals) + i, 0 ≤ j, 0 ≤ i, i ≤ j, j ≤ n→i(# vals) + i ⊦ sk = _type, sk = _length, (r -- sk ∈ putArray(vals, r, i, addarray(r, ty, j, st)) → r -- sk ∈ st) ∧ st[r -- sk] = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st)[r -- sk])

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 205:
⊦ 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 210:
is_fskey(sk), 0 ≤ n→i(# vals) + i, 0 ≤ j, 0 ≤ i, i ≤ j, j ≤ n→i(# vals) + i ⊦ sk = _type, sk = _length, ( r -- sk ∈ addarray(r, ty, j, st) ∨ (r -- sk) .ref = r ∧ is_indexkey((r -- sk) .key) ∧ i ≤ (r -- sk) .key .index ∧ (r -- sk) .key .index < i + n→i(# vals) → r -- sk ∈ st) ∧ st[r -- sk] = (addarray(r, ty, mklist(initial_value(ty), i→n(i)) + vals, st)[r -- sk])

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, j, st), i, r, vals, r -- sk] ,F ,storearrays , ,putArray-in ]

Additional information

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