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Proof step 44 from proof subset-size (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 41:
⊦ st1 ⊆ st2 ↔ (∀ rk. rk ∈ st1 → rk ∈ st2 ∧ st1[rk] = st2[rk])
Proof step 43:
r ∈ st0 .refs, # st0 .refs ≤ # st1 .refs, ∀ rk. rk ∈ st1 -- (r -- sk) → rk ∈ st0 -- (r -- sk) ∧ (st1 -- (r -- sk))[rk] = (st0 -- (r -- sk))[rk], r -- sk ∈ st0, r ∈ st0, rk ∈ st1, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ st0 = ∅, r ∈ 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 -- (r -- sk)] ,F ,store ,javastore ,Subset ]

Additional information

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