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Proof step 5 from proof addbar-in-01 (conclusion)

⊦ r ∈ addbytearray(r0, ints, st) ↔ r = r0 ∨ r ∈ st

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ addbytearray(r, ints, st) = addarray(r, byte_type, mkjavavalues(is2bs(ints)), st)
Proof step 2:
⊦ r ≠ r0 → (r ∈ addarray(r0, ty0, vals, st) ↔ r ∈ st)
Proof step 3:
⊦ r ∈ addarray(r, ty0, vals, st)
Proof step 4:
⊦ r ∈ st → r ∈ addarray(r0, ty0, vals, st)

Rule argument

[crewrite-arg: addbytearray-ints, F, [st, ints, r] → [st, ints, r0]]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, ints, r0)]]