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Proof step 1 from proof diff-empty (conclusion)
⊦ s1 \ s2 = ∅ ∧ a ∈ s1 → a ∈ s2
System applied rule normalize
And obtained the following new goals
Proof step 2
:
s1 \ s2 = ∅, a ∈ s1 ⊦ a ∈ s2
Rule argument
Additional information