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Proof step 25 from proof subset-size (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ¬ r ∈ rset → # (rset ++ r) = # rset + 1
Proof step 23:
¬ ¬ r ∈ st0 .refs, r ∈ st0, r -- sk ∈ st0, st1 -- (r -- sk) ⊆ st0 -- (r -- sk), # st0 .refs < # st1 .refs, r ∈ st1 .refs, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st, # st1 .refs ≤ # (st0 .refs ++ r) ∨ ¬ st1 ⊆ st0[r -- sk, st1[r -- sk]] ⊦ st1 = ∅
Proof step 24:
⊦ ¬ r ∈ st .refs → ¬ r ∈ st

Rule argument

[speclemmaarg: ,¬ r ∈ rset → # (rset ++ r) = # rset + 1 ,[r, rset] → [r, st0 .refs] ,F ,set-basic ,references ,Size-insert ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, st0 .refs)]]