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Proof step 74 from proof junkrefs-union (conclusion)

⊦ (∃ st0. st1 = st0 ∪ st) ∧ junkrefs(str, refs, ty, st1) → junkrefs(str, refs, ty, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]