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Proof step 18 from proof in-add (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 17:
∀ 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

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : sk ≠ sk0 ⊦ r -- sk ∈ st[r -- sk0, val] ↔ r -- sk ∈ st], [CSIMPSEQ : ⊦ _length ≠ i '], [CSIMPSEQ : ⊦ n -1 = n - 1])]]