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Proof step 30 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), vals) + vals[i→n(i) + i→n(i0)], (~1) = i0, vals = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 29:
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, i + i0 < n→i(# vals), 0 < i0, 0 ≤ i + i0, i→n(i) + i→n(i0) < # vals, ∀ 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)) ⊦ vals[i→n(i + i0)] = vals[i→n(i) + i→n(i0)], vals = [], (~1) = i0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # vals = 0 ↔ vals = []], [CSIMPFORWARD : ⊦ m < n → n ≠ 0], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ (vals + vals0) + vals1 = vals + vals0 + vals1], [CSIMPSEQ : ⊦ val ' = val0 ' ↔ val = val0], [CSIMPSEQ : ⊦ vals + vals0 = vals + vals1 ↔ vals0 = vals1], [CSIMPSEQ : ⊦ vals + val = vals + val '])]]