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Proof step 15 from proof union-subset-left (conclusion)
⊦ st .refs ∩ st0 .refs = ∅ → st ⊆ st ∪ st0
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 14
:
st .refs ∩ st0 .refs = ∅ ⊦ st ⊆ st ∪ st0
Rule argument
[emptyarg]
Additional information
[[CONCRETESIMPRULES : ()]]