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Proof step 1 from proof subseteq-lebrefs-subseteq (conclusion)
⊦ q0 ⊆ q1 → q0.lebrefs ⊆ q1.lebrefs
System applied rule normalize
And obtained the following new goals
Proof step 2
:
q0 ⊆ q1 ⊦ q0.lebrefs ⊆ q1.lebrefs
Rule argument
Additional information