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Proof step 7 from proof isarray-02 (conclusion)

⊦ i ≤ j → (is_arrayref(r, i, st[r0 -- j ', val]) ↔ is_arrayref(r, i, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ is_arrayref(r, i, st) ↔ (∀ j. 0 ≤ j ∧ j < i → r -- j ' ∈ st)
Proof step 6:
i ≤ j ⊦ (∀ j0. ¬ j0 < 0 ∧ ¬ i ≤ j0 → r -- j0 ' ∈ st[r0 -- j ', val]) ↔ (∀ j. 0 ≤ j ∧ j < i → r -- j ' ∈ st)

Rule argument

[emptyarg]

Additional information

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