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Proof step 24 from proof stsubset-new (previous step)

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

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 23:
∀ rk. rk ∈ st → rk .ref ≠ r ⊦ (∀ rk. rk ∈ st → rk ∈ st0[r -- sk, val] ∧ st[rk] = (st0[r -- sk, val][rk])) ↔ st ⊆ st0, r = jvmref

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0[r -- sk, val], st)]]