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Proof step 16 from proof get-add-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ (addarray(r, vals, j, st[r -- _length, val])[r -- i ']) = (addarray(r, vals, j, st)[r -- i '])
Proof step 6:
⊦ (addarray(r, vals, j, st[r -- _type, val])[r -- i ']) = (addarray(r, vals, j, st)[r -- i '])
Proof step 15:
i < n→i(# vals), 0 ≤ i, ∀ 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 = [], (addarray(r, vals, 0, st)[r -- i ']) ' = sublist(i→n(i), 1, vals)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]