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Proof step 5 from proof is_arrayref-03 (previous step)

j0 < i, is_arrayref(r, i, st), 0 ≤ j0, 0 < i ⊦ j0 = j, r -- j0 ' ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ is_arrayref(r, i, st) ↔ (∀ j. 0 ≤ j ∧ j < i → r -- j ' ∈ st)
Proof step 4:
j0 < i, ∀ j. 0 ≤ j ∧ j < i → r -- j ' ∈ st, 0 ≤ j0, 0 < i ⊦ j0 = j, r -- j0 ' ∈ st

Rule argument

[crewrite-arg: is_arrayref, F, [st, i, r] → [st, i, r]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, i, r)]]