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Proof step 82 from proof union-addarray-rew (previous step)

newref(r, st), newref(r, st0) ⊦ r = jvmref, (r0 -- sk ∈ addarray(r, ty, vals, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addarray(r, ty, vals, st0)) ∧ (addarray(r, ty, vals, st ∪ st0)[r0 -- sk]) = (st ∪ addarray(r, ty, vals, st0))[r0 -- sk]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ r ≠ r0 → (addarray(r, ty, vals, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 74:
¬ r ≠ r0, newref(r, st), newref(r, st0) ⊦ r = jvmref, (r0 -- sk ∈ addarray(r, ty, vals, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addarray(r, ty, vals, st0)) ∧ (addarray(r, ty, vals, st ∪ st0)[r0 -- sk]) = (st ∪ addarray(r, ty, vals, st0))[r0 -- sk]
Proof step 81:
r ≠ r0, newref(r, st), newref(r, st0) ⊦ r = jvmref, (r0 -- sk ∈ addarray(r, ty, vals, st ∪ st0) ↔ r0 -- sk ∈ st ∨ r0 -- sk ∈ addarray(r, ty, vals, st0)) ∧ (st ∪ st0)[r0 -- sk] = (st ∪ addarray(r, ty, vals, st0))[r0 -- sk]

Rule argument

[crewrite-arg: lookup-1init , F , [sk, r0, st, vals, ty, r] → [sk, r0, st ∪ st0, vals, ty, r] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sk, r0, st ∪ st0, vals, ty, r)]]