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Proof step 42 from proof get-add-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 41:
0 ≤ j, 0 ≤ i, i + j ≤ n→i(# vals), ∀ ty, st, r, vals, i, j0. i→n(j0) < i→n(j) → ( 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)) ⊦ getarray(r, i, j, addarray(r, vals, 0, st[r -- _type, typeval(mkarraytype(ty))][r -- _length, intval(n→i(# vals))])) = sublist(i→n(i), i→n(j), vals)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]