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Proof step 31 from proof refs-equal (previous step)

# st .refs = # st0 .refs, st ⊆ st0, r ∈ st0, r0 ∈ st .refs, r0 ∈ st0 .refs, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ st = ∅, st0 = ∅, r = r0, r ∈ st, st -- r0 ⊆ st0 -- r0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 20:
⊦ st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk])
Proof step 30:
# st .refs = # st0 .refs, st ⊆ st0, r ∈ st0, r0 ∈ st .refs, r0 ∈ st0 .refs, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ st = ∅, st0 = ∅, r = r0, r ∈ st, ∀ rk. rk ∈ st -- r0 → rk ∈ st0 -- r0 ∧ (st -- r0)[rk] = (st0 -- r0)[rk]

Rule argument

[speclemmaarg: ,st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk]) ,[st2, st1] → [st0 -- r0, st -- r0] ,F ,store ,javastore ,Subset ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0 -- r0, st -- r0)]]