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Proof step 27 from proof get-add-02 (previous step)

¬ i→n(i) + i→n(i0) < # vals, 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, ∀ 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 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 26:
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 = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ m < n ↔ n ≤ m], [CSIMPSEQ : ⊦ i + j = j + i])]]