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Proof step 17 from proof in-add (previous step)

∀ 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(# vals - 1) ', vals .last]) ↔ ¬ j < 0 ∧ ¬ n→i(# vals) ≤ j ∨ r -- j ' ∈ st

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 16:
# 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

Rule argument

[speclemmaarg: n ≠ 0 ,m = n - 1 ↔ n = m + 1 ,[n, m] → [# vals, m] ,T ,nat , ,elim-pred-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # vals = 0 ↔ vals = []], [CSIMPELIM : n ≠ 0 ⊦ m = n - 1 ↔ n = m + 1])], [EXTRATERMS : (# vals, m)]]