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Proof step 18 from proof subset-size (previous step)

st1 ⊆ st0 -- (r -- sk), # st0 .refs < # st1 .refs, r ∈ st1 .refs, r ∈ st0 .refs, r ∈ st0, r -- sk ∈ st0, rk ∈ st1, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, st1 = ∅, r -- sk ∈ st1, rk ∈ st0 ∧ st1[rk] = st0[rk]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 11:
⊦ st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk])
Proof step 17:
∀ rk. rk ∈ st1 → rk ∈ st0 -- (r -- sk) ∧ st1[rk] = (st0 -- (r -- sk))[rk], # st0 .refs < # st1 .refs, r ∈ st1 .refs, r ∈ st0 .refs, r ∈ st0, r -- sk ∈ st0, rk ∈ st1, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, st1 = ∅, r -- sk ∈ st1, rk ∈ st0 ∧ st1[rk] = st0[rk]

Rule argument

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

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st0 -- (r -- sk), st1)]]