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Proof step 2 from proof elems-in-diff (
previous step
)
n ∈ nx0 ⊦ n ∈ elems nx0 \ elems nx1, n ∈ nx1
Heuristic simplifier applied rule simplifier
And obtained the following new goals
Proof step 3
:
n ∈ nx0 ⊦ n ∈ nx1, n ∈ elems nx0 \ elems nx1
Rule argument
Additional information