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Proof step 6 from proof subset-addarray-01 (previous step)

st[r0 -- sk] = (addarray(r, ty, vals, st0)[r0 -- sk]), newref(r, st), newref(r, st0), r0 -- sk ∈ st, r0 ∈ st, r0 -- sk ∈ addarray(r, ty, vals, st0), r0 ∈ addarray(r, ty, vals, st0), ∀ rk. rk ∈ st → rk ∈ addarray(r, ty, vals, st0) ∧ st[rk] = (addarray(r, ty, vals, st0)[rk]) ⊦ r = jvmref, r0 -- sk ∈ st0 ∧ (addarray(r, ty, vals, st0)[r0 -- sk]) = st0[r0 -- sk]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ r ≠ r0 → (r -- sk ∈ addarray(r0, ty, vals, st) ↔ r -- sk ∈ st)
Proof step 4:
⊦ r ≠ r0 → (addarray(r, ty, vals, st)[r0 -- sk]) = st[r0 -- sk]
Proof step 5:
⊦ newref(r, st) ∧ r0 ∈ st → (addarray(r, ty, vals, st)[r0 -- sk]) = st[r0 -- sk]

Rule argument

[crewrite-arg: in-init-01 , F , [st, vals, ty, r0, sk, r] → [st0, vals, ty, r, sk, r0] ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ r ∈ st → ¬ newref(r, st)], [CSIMPSEQ : ⊦ newref(r, st) ∧ r0 ∈ st → r ≠ r0], [CSIMPFORWARD : ⊦ r -- sk ∈ st → r ∈ st])], [EXTRATERMS : (st0, vals, ty, r, sk, r0)]]