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Proof step 22 from proof stsubset-new (previous step)

∀ rk. rk ∈ st → rk .ref ≠ r ⊦ r = jvmref, (∀ rk. rk ∈ st → st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ rk ∈ st0[r -- sk, val]) ↔ st ⊆ st0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk])
Proof step 21:
∀ rk. rk ∈ st → rk .ref ≠ r ⊦ r = jvmref, (∀ rk. rk ∈ st → st[rk] = ((rk = r -- sk ⊃ val; st0[rk])) ∧ rk ∈ st0[r -- sk, val]) ↔ (∀ rk. rk ∈ st → rk ∈ st0 ∧ st[rk] = st0[rk])

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)]]