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Proof step 13 from proof is_obj-union-disjoint (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ is_obj(str, r, st) ↔ class+∃(str, st[r -- _type] .type .class) ∧ ¬ is_abstract(st[r -- _type] .type .class) ∧ is_obj(r, st)
Proof step 12:
st1 .refs ∩ st .refs = ∅, r ∈ st1, class+∃(str, st1[r -- _type] .type .class), is_obj(r, st1) ⊦ class+∃(str, (st1 ∪ st)[r -- _type] .type .class) ∧ ¬ is_abstract((st1 ∪ st)[r -- _type] .type .class) ∧ is_obj(r, st1 ∪ st), is_abstract(st1[r -- _type] .type .class)

Rule argument

[crewrite-arg: is-real-object-def , F , [st, r, str] → [st1 ∪ st, r, str] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (st1 ∪ st, r, str)]]