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Proof step 14 from proof is_obj-union-disjoint (conclusion)

⊦ st1 .refs ∩ st .refs = ∅ ∧ r ∈ st1 ∧ is_obj(str, r, st1) → is_obj(str, r, st1 ∪ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ 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 13:
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)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]