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Proof step 5 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, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals) ≤ j ∨ r -- j ' ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
∀ st, ty, j, r, vals0. # vals0 < # [] → (true → (r -- j ' ∈ addarray(r, ty, vals0, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals0) ≤ j ∨ r -- j ' ∈ st)) ⊦ r -- j ' ∈ addarray(r, ty, [], st) ↔ r -- j ' ∈ st ∨ ¬ j < 0 ∧ ¬ 0 ≤ j

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ # [] = 0])]]