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Proof step 4 from proof isarray-03 (conclusion)

⊦ is_arrayref(r, 1, st) ↔ r -- 0 ' ∈ 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 3:
⊦ (∀ j. ¬ j < 0 ∧ ¬ 0 < j → r -- j ' ∈ st) ↔ r -- 0 ' ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ i < 1 ↔ i ≤ 0], [CSIMPSEQ : ⊦ ¬ i ≤ j ↔ j < i])]]