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Proof step 50 from proof subset-size (previous step)

# (st0 .refs -- r) < # st1 .refs, # st0 .refs ≤ # st1 .refs, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), r -- sk ∈ st0, r ∈ st0, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ r ∈ st1 .refs, st1 ⊆ st0 -- r

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 34:
⊦ r ∈ rset → # (rset -- r) = # rset - 1
Proof step 35:
⊦ ¬ r ∈ st .refs → ¬ r -- sk ∈ st
Proof step 49:
r ∈ st0 .refs, # st0 .refs - 1 < # st1 .refs, # st0 .refs ≤ # st1 .refs, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), r -- sk ∈ st0, r ∈ st0, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦ r ∈ st1 .refs, st1 ⊆ st0 -- r

Rule argument

[speclemmaarg: ,r ∈ rset → # (rset -- r) = # rset - 1 ,[r, rset] → [r, st0 .refs] ,F ,set-union ,references ,del-size ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ r ∈ rset → rset -- r = rset], [CSIMPSEQ : ⊦ m < n → m ≤ n])], [EXTRATERMS : (r, st0 .refs)]]