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Proof step 5 from proof union-at-right-02 (conclusion)

⊦ ¬ rk ∈ st → (st ∪ st0)[rk] = st0[rk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (st ∪ st0)[rk] = st0[rk], rk ∈ st

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]