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Proof step 8 from proof add-at-neg (previous step)

# vals = m + 1, i < 0, ∀ st, ty, r, i, vals0. # vals0 < m + 1 → (i < 0 → (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 if-then-else split applied rule case distinction

And obtained the following new goals

Proof step 7:
i ≠ n→i(m), # vals = m + 1, i < 0, ∀ st, ty, r, i, vals0. # vals0 < m + 1 → (i < 0 → (addarray(r, ty, vals0, st)[r -- i ']) = st[r -- i ']) ⊦ (addarray(r, ty, vals .butlast, st)[r -- i ']) = st[r -- i '], vals = []

Rule argument

[CASEDARG : pos: right 2, ()]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ n→i(n) < 0])]]