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Proof step 8 from proof union-subset-left (previous step)

st .refs ∩ st0 .refs = ∅, r -- sk ∈ st0, r -- sk ∈ st, r ∈ st, r ∈ st0 ⊦ st[r -- sk] = st0[r -- sk]

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ rset ∩ rset0 = ∅ ↔ (∀ r. ¬ r ∈ rset ∨ ¬ r ∈ rset0)
Proof step 7:
∀ r. ¬ r ∈ st .refs ∨ ¬ r ∈ st0 .refs, r -- sk ∈ st0, r -- sk ∈ st, r ∈ st, r ∈ st0 ⊦ st[r -- sk] = st0[r -- sk]

Rule argument

[speclemmaarg: ,rset ∩ rset0 = ∅ ↔ (∀ r. ¬ r ∈ rset ∨ ¬ r ∈ rset0) ,[rset0, rset] → [st0 .refs, st .refs] ,F ,set-union ,references ,int-empty-01 ]

Additional information

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