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Proof step 22 from proof junkref-union-disjoint (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 21:
st1 .refs ∩ st .refs = ∅, junkref(str, r, st1 ∪ st), r ∈ st1 ⊦ junkref(str, r, st1)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]