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Proof step 6 from proof union-at-left (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ (st ∪ st0)[rk] = st[rk], rk ∈ st0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]