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Proof step 16 from proof in-add (previous step)

# vals = m + 1 ∧ # vals ≠ 0, ∀ st, ty, j, r, vals0. # vals0 < # vals → (true → (r -- j ' ∈ addarray(r, ty, vals0, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals0) ≤ j ∨ r -- j ' ∈ st)) ⊦ vals = [], r -- j ' ∈ (addarray(r, ty, vals .butlast, st)[r -- n→i(m) ', vals .last]) ↔ ¬ j < 0 ∧ ¬ n→i(# vals) ≤ j ∨ r -- j ' ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
# vals = m + 1, ∀ st, ty, j, r, vals0. # vals0 < m + 1 → (true → (r -- j ' ∈ addarray(r, ty, vals0, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals0) ≤ j ∨ r -- j ' ∈ st)) ⊦ vals = [], r -- j ' ∈ (addarray(r, ty, vals .butlast, st)[r -- n→i(m) ', vals .last]) ↔ r -- j ' ∈ st ∨ ¬ j < 0 ∧ ¬ n→i(m) < j

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([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 = []])]]