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Proof step 26 from proof get-add-02 (previous step)

getarray(r, i, i0, addarray(r, vals, 0, st[r -- _type, typeval(mkarraytype(ty))][r -- _length, intval(n→i(# vals))])) = sublist(i→n(i), i→n(i0), vals), 0 ≤ i + i0, 0 < i0, i + i0 < n→i(# vals), 0 ≤ i, # vals ≤ i→n(i) + i→n(i0), ∀ ty, st, r, vals, i, j0. i→n(j0) < i→n(i0 + 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)) ⊦ sublist(i→n(i), i→n(i0), vals) + vals[i→n(i + i0)] ' = sublist(i→n(i), i→n(i0) + 1, vals), (~1) = i0, vals = []

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 23:
0 ≤ i ⊦ i < n→i(n) ↔ i→n(i) < n
Proof step 25:
0 ≤ i + i0, getarray(r, i, i0, addarray(r, vals, 0, st[r -- _type, typeval(mkarraytype(ty))][r -- _length, intval(n→i(# vals))])) = sublist(i→n(i), i→n(i0), vals), 0 ≤ i + i0, 0 < i0, i→n(i + i0) < # vals, 0 ≤ i, # vals ≤ i→n(i) + i→n(i0), ∀ ty, st, r, vals, i, j0. i→n(j0) < i→n(i0 + 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)) ⊦ sublist(i→n(i), i→n(i0), vals) + vals[i→n(i + i0)] ' = sublist(i→n(i), i→n(i0) + 1, vals), (~1) = i0, vals = []

Rule argument

[speclemmaarg: 0 ≤ i ,((* i)) < n→i(n) ↔ i→n(i) < n ,[n, i] → [# vals, i + i0] ,F ,intnat , ,num-n2i-ls-02 ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i + j = j + i])], [EXTRATERMS : (# vals, i + i0)]]