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Proof step 8 from proof junkref-union-disjoint (previous step)

st1 .refs ∩ st .refs = ∅, r ∈ st, junkrefs(str, r ' + refs0, st[r -- _type] .type, st1 ∪ st), r ∈ st1, r -- _type ∈ st, r ∈ st .refs ⊦ st = ∅, junkrefs(str, r ' + refs0, st1[r -- _type] .type, st1), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st1[refs .first -- _type] .type, st1)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ r ∈ st ↔ r ∈ st .refs
Proof step 7:
st1 .refs ∩ st .refs = ∅, r ∈ st .refs, junkrefs(str, r ' + refs0, st[r -- _type] .type, st1 ∪ st), r ∈ st1, r -- _type ∈ st, r ∈ st .refs ⊦ st = ∅, junkrefs(str, r ' + refs0, st1[r -- _type] .type, st1), ∃ refs. refs ≠ [] ∧ refs .first = r ∧ junkrefs(str, refs, st1[refs .first -- _type] .type, st1)

Rule argument

[speclemmaarg: ,r ∈ st ↔ r ∈ st .refs ,[st, r] → [st, r] ,F ,storefunsx , ,instore-inrefs ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st, r)]]