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Proof step 29 from proof subset-delete (previous step)

⊦ st ⊆ st0 -- r ↔ st ⊆ st0, r ∈ 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 28:
⊦ (∀ rk. rk ∈ st → rk ∈ st0 -- r ∧ st[rk] = (st0 -- r)[rk]) ↔ st ⊆ st0, r ∈ st

Rule argument

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

Additional information

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