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Proof step 6 from proof is_arrayref-01 (previous step)

∀ j. 0 ≤ j ∧ j < i → r -- j ' ∈ st ⊦ ∀ j. ¬ j < 0 ∧ ¬ i ≤ j → r -- j ' ∈ st[_out, val]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
j < i, 0 < i, 0 ≤ j, ∀ j0. ¬ j0 < 0 ∧ ¬ i ≤ j0 → r -- j0 ' ∈ st ⊦ r -- j ' ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPFORWARD : ⊦ i < j ∧ 0 ≤ i → 0 < j], [CSIMPFORWARD : ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j], [CSIMPSEQ : ⊦ ¬ i < j ↔ j ≤ i], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i], [CSIMPSEQ : ⊦ _out ≠ r -- i '], [CSIMPSEQ : ⊦ rk ∈ st[rk0, val] ↔ rk = rk0 ∨ rk ∈ st])]]