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Proof step 11 from proof add-at-neg (previous step)

vals ≠ [], i < 0, ∀ st, ty, r, i, vals0. # vals0 < # vals → (i < 0 → (addarray(r, ty, vals0, st)[r -- i ']) = st[r -- i ']) ⊦ (addarray(r, ty, vals .butlast, st)[r -- n→i((# vals) -1) ', vals .last][r -- _length, intval(n→i(# vals))][r -- i ']) = st[r -- i ']

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
i < 0, ∀ st, ty, r, i, vals0. # vals0 < # vals → (i < 0 → (addarray(r, ty, vals0, st)[r -- i ']) = st[r -- i ']) ⊦ vals = [], ((i = n→i(# vals - 1) ⊃ vals .last; addarray(r, ty, vals .butlast, st)[r -- i '])) = st[r -- i ']

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i ' = i0 ' ↔ i = i0], [CSIMPSEQ : ⊦ _length ≠ i '], [CSIMPSEQ : ⊦ r -- sk = r0 -- sk0 ↔ ¬ ¬ (r = r0 ∧ sk = sk0)], [CSIMPSEQ : ⊦ (st[rk0, val][rk]) = ((rk = rk0 ⊃ val; st[rk]))], [CSIMPSEQ : ⊦ n -1 = n - 1])]]