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Proof step 10 from proof is_array-init-01 (previous step)

⊦ r = r0, (∀ j0. 0 ≤ j0 ∧ j0 < i → r -- j0 ' ∈ addarray(r0, vals, j, 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, vals, j, st) ↔ r -- sk ∈ st)
Proof step 9:
⊦ r = r0, (∀ j0. ¬ j0 < 0 ∧ ¬ i ≤ j0 → r -- j0 ' ∈ st) ↔ is_arrayref(r, i, st)

Rule argument

[emptyarg]

Additional information

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