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Proof step 12 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) ⊦ 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ is_obj(r, st1) → is_obj(r, st1 ∪ st)
Proof step 11:
st1 .refs ∩ st .refs = ∅, r ∈ st1, class+∃(str, st1[r -- _type] .type .class), is_obj(r, st1) ⊦ is_abstract(st1[r -- _type] .type .class), class+∃(str, (st1 ∪ st)[r -- _type] .type .class) ∧ ¬ is_abstract((st1 ∪ st)[r -- _type] .type .class)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ rset ∩ rset0 = rset0 ∩ rset], [CSIMPSEQ : ⊦ (rset ∩ rset0) ∩ rset1 = rset ∩ rset0 ∩ rset1])]]