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Proof step 11 from proof putarray-subset-new-01 (previous step)

∀ rk. rk ∈ st → rk ∈ putArray(vals, r, i, st0) ∧ st[rk] = (putArray(vals, r, i, st0)[rk]) ⊦ st ⊆ st0, r ∈ st

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 10:
∀ rk. rk ∈ st → rk ∈ putArray(vals, r, i, st0) ∧ st[rk] = (putArray(vals, r, i, st0)[rk]) ⊦ ∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], r ∈ st

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0, st)]]