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Proof step 10 from proof junkref-union-disjoint (previous step)

st1 .refs ∩ st .refs = ∅, r -- _type ∈ st, r ∈ st1, junkrefs(str, r ' + refs0, st[r -- _type] .type, st1 ∪ st), r ∈ 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 4:
⊦ ¬ r ∈ st .refs → ¬ r -- sk ∈ st
Proof step 9:
¬ ¬ r ∈ st .refs, st1 .refs ∩ st .refs = ∅, r -- _type ∈ st, r ∈ st1, junkrefs(str, r ' + refs0, st[r -- _type] .type, st1 ∪ st), r ∈ 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 .refs → ¬ r -- sk ∈ st ,[st, sk, r] → [st, _type, r] ,F ,storefunsx , ,refs-notin ]

Additional information

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