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Proof step 13 from proof get-add-02 (previous step)

0 ≤ i, i < n→i(# vals), ∀ ty, st, r, vals, i, j0. i→n(j0) < i→n(1) → ( i + j0 ≤ n→i(# vals) ∧ 0 ≤ i ∧ 0 ≤ j0 → getarray(r, i, j0, addarray(r, vals, 0, st[r -- _type, typeval(mkarraytype(ty))][r -- _length, intval(n→i(# vals))])) = sublist(i→n(i), i→n(j0), vals)) ⊦ vals[i→n(i)] ' = sublist(i→n(i), 1, vals), vals = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 9:
⊦ m < # vals → sublist(m, 1, vals) = vals[m] '
Proof step 12:
¬ i→n(i) < # vals, 0 ≤ i, i < n→i(# vals), ∀ ty, st, r, vals, i, j0. i→n(j0) < i→n(1) → ( i + j0 ≤ n→i(# vals) ∧ 0 ≤ i ∧ 0 ≤ j0 → getarray(r, i, j0, addarray(r, vals, 0, st[r -- _type, typeval(mkarraytype(ty))][r -- _length, intval(n→i(# vals))])) = sublist(i→n(i), i→n(j0), vals)) ⊦ vals[i→n(i)] ' = sublist(i→n(i), 1, vals), vals = []

Rule argument

[speclemmaarg: ,m < # vals → sublist(m, 1, vals) = vals[m] ' ,[vals, m] → [vals, i→n(i)] ,F ,list-del ,javavalues ,sublist-one ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (vals, i→n(i))]]