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Proof step 44 from proof refs-equal (previous step)

# st .refs = # st0 .refs, st ⊆ st0, r ∈ st0, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r ∈ st

User applied rule insert spec-lemma

And obtained the following new goals

Proof step 1:
⊦ rset = ∅ ∨ (∃ r. r ∈ rset)
Proof step 43:
st .refs = ∅ ∨ (∃ r. r ∈ st .refs), # st .refs = # st0 .refs, st ⊆ st0, r ∈ st0, ∀ r, st0, st1. # st1 .refs < # st .refs → (# st1 .refs = # st0 .refs ∧ r ∈ st0 ∧ st1 ⊆ st0 → r ∈ st1) ⊦ r ∈ st

Rule argument

[speclemmaarg: ,rset = ∅ ∨ (∃ r. r ∈ rset) ,[rset] → [st .refs] ,F ,set-basic ,references ,sel ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st .refs)]]