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Proof step 12 from proof subset-putarray-addarray-newref (previous step)

newref(r, st), rk ∈ st, rk ∈ putArray(vals0, r, i, addarray(r, ty, vals, st)) ⊦ st[rk] = (putArray(vals0, r, i, addarray(r, ty, vals, st))[rk]), r = jvmref

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ rk ∈ st ∧ st ⊆ st1 → st1[rk] = st[rk]
Proof step 5:
⊦ rk ∈ st → rk ∈ addarray(r, ty, vals, st)
Proof step 11:
rk ∈ putArray(vals0, r, i, addarray(r, ty, vals, st)) ∧ putArray(vals0, r, i, addarray(r, ty, vals, st)) ⊆ putArray(vals0, r, i, addarray(r, ty, vals, st)), newref(r, st), rk ∈ st, rk ∈ putArray(vals0, r, i, addarray(r, ty, vals, st)) ⊦ st[rk] = (putArray(vals0, r, i, addarray(r, ty, vals, st))[rk]), r = jvmref

Rule argument

[speclemmaarg: ,rk ∈ st ∧ st ⊆ st1 → st1[rk] = st[rk] ,[rk, st, st1] → [rk, putArray(vals0, r, i, addarray(r, ty, vals, st)), putArray(vals0, r, i, addarray(r, ty, vals, st))] ,F ,store ,javastore ,superset-elem-eq ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rk ∈ putArray(vals, r, i, st) ↔ rk ∈ st ∨ rk .ref = r ∧ is_indexkey(rk .key) ∧ i ≤ rk .key .index ∧ rk .key .index < i + n→i(# vals)], [CSIMPSEQ : ⊦ st ⊆ st])], [EXTRATERMS : (rk, putArray(vals0, r, i, addarray(r, ty, vals, st)), putArray(vals0, r, i, addarray(r, ty, vals, st)))]]