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Proof step 3 from proof isarray-union-disjoint (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
st1 .refs ∩ st .refs = ∅, is_array(r, st1), r ∈ st1 ⊦ is_array(r, st1 ∪ st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]