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Proof step 22 from proof in-add (conclusion)

⊦ r -- j ' ∈ addarray(r, ty, vals, st) ↔ r -- j ' ∈ st ∨ 0 ≤ j ∧ j < n→i(# vals)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
⊦ r -- j ' ∈ addarray(r, ty, vals, st) ↔ ¬ j < 0 ∧ ¬ n→i(# vals) ≤ j ∨ r -- j ' ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i])]]