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Proof step 9 from proof subset-addarray-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk])
Proof step 8:
newref(r, st), newref(r, st0), ∀ rk. rk ∈ st → rk ∈ addarray(r, ty, vals, st0) ∧ st[rk] = (addarray(r, ty, vals, st0)[rk]), r0 -- sk ∈ st, r0 ∈ st ⊦ r = jvmref, r0 -- sk ∈ st0 ∧ st[r0 -- sk] = st0[r0 -- sk]

Rule argument

[speclemmaarg: ,st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk]) ,[st2, st1] → [addarray(r, ty, vals, st0), st] ,F ,store ,javastore ,Subset ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addarray(r, ty, vals, st0), st)]]