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Proof step 9 from proof is_obj-union-disjoint (previous step)

st1 .refs ∩ st .refs = ∅, r -- _type ∈ st, r ∈ st1, class+∃(str, st1[r -- _type] .type .class), is_obj(r, st1), r ∈ st ⊦ is_abstract(st1[r -- _type] .type .class), class+∃(str, st[r -- _type] .type .class) ∧ ¬ is_abstract(st[r -- _type] .type .class)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ ¬ r ∈ st .refs → ¬ r -- sk ∈ st
Proof step 8:
¬ ¬ r ∈ st .refs, st1 .refs ∩ st .refs = ∅, r -- _type ∈ st, r ∈ st1, class+∃(str, st1[r -- _type] .type .class), is_obj(r, st1), r ∈ st ⊦ is_abstract(st1[r -- _type] .type .class), class+∃(str, st[r -- _type] .type .class) ∧ ¬ is_abstract(st[r -- _type] .type .class)

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)]]