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Proof step 56 from proof subset-size (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ ¬ r ∈ rset → # (rset ++ r) = # rset + 1
Proof step 30:
¬ ¬ r ∈ st1 .refs, st1[r -- sk, val] ⊆ st0, # st0 .refs < # (st1 .refs ++ r), ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦
Proof step 55:
¬ r ∈ st1 .refs, st1[r -- sk, val] ⊆ st0, # st0 .refs < # st1 .refs + 1, ∀ st. # st1 .refs ≤ # st .refs ∨ ¬ st1 ⊆ st ⊦

Rule argument

[speclemmaarg: ,¬ r ∈ rset → # (rset ++ r) = # rset + 1 ,[r, rset] → [r, st1 .refs] ,F ,set-basic ,references ,Size-insert ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (r, st1 .refs)]]