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Proof step 3 from proof subset-eqval (previous step)

st ⊆ st0, rk ∈ st ⊦ st0[rk] .intval = st[rk] .intval

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 2:
∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk], rk ∈ st ⊦ st0[rk] .intval = st[rk] .intval

Rule argument

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

Additional information

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