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Proof step 7 from proof is_array-init (previous step)

⊦ r = r0, (∀ j. 0 ≤ j ∧ j < i → r -- j ' ∈ addarray(r0, ty0, vals, st)) ↔ is_arrayref(r, i, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, vals, st) ↔ r -- sk ∈ st)
Proof step 6:
⊦ r = r0, (∀ j. ¬ j < 0 ∧ ¬ i ≤ j → r -- j ' ∈ st) ↔ is_arrayref(r, i, st)

Rule argument

[emptyarg]

Additional information

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