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Proof step 9 from proof add-at-gr (previous step)

# vals = m + 1 ∧ # vals ≠ 0, n→i(# vals) ≤ i, ∀ st, ty, r, i, vals0. # vals0 < # vals → (n→i(# vals0) ≤ i → (addarray(r, ty, vals0, st)[r -- i ']) = st[r -- i ']) ⊦ vals = [], ((i = n→i(m) ⊃ vals .last; addarray(r, ty, vals .butlast, st)[r -- i '])) = st[r -- i ']

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
# vals = m + 1, n→i(m) < i, 0 < i, ∀ st, ty, r, i, vals0. # vals0 < m + 1 → (n→i(# vals0) ≤ i → (addarray(r, ty, vals0, st)[r -- i ']) = st[r -- i ']) ⊦ ((i = n→i(m) ⊃ vals .last; addarray(r, ty, vals .butlast, st)[r -- i '])) = st[r -- i '], vals = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ n→i(m) < j → 0 < j], [CSIMPSEQ : ⊦ i + j = j + i], [CSIMPSEQ : ⊦ i + j + i0 = (i + j) + i0], [CSIMPSEQ : ⊦ i + 1 ≤ j ↔ i < j], [CSIMPSEQ : ⊦ m + n = n + m], [CSIMPSEQ : ⊦ (m + n0) + n1 = m + n0 + n1], [CSIMPSEQ : ⊦ n→i(m + n) = n→i(m) + n→i(n)], [CSIMPSEQ : ⊦ # vals = 0 ↔ vals = []])]]