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Proof step 20 from proof addarrays-subset (conclusion)

⊦ is_newref_list(refs, st) ∧ 0 ≤ i → st ⊆ addarrays(refs, ty, ints, i, st)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk])
Proof step 19:
⊦ is_newref_list(refs, st) ∧ 0 ≤ i → (∀ rk. rk ∈ st → rk ∈ addarrays(refs, ty, ints, i, st) ∧ st[rk] = (addarrays(refs, ty, ints, i, st)[rk]))

Rule argument

[speclemmaarg: ,st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk]) ,[st2, st1] → [addarrays(refs, ty, ints, i, st), st] ,F ,store ,javastore ,Subset ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (addarrays(refs, ty, ints, i, st), st)]]