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Proof step 1 from proof ebm-order-trans (conclusion)
⊦ avols ⊆ avols0 ∧ avols0 ⊆ avols1 → avols ⊆ avols1
System applied rule normalize
And obtained the following new goals
Proof step 2
:
avols ⊆ avols0, avols0 ⊆ avols1 ⊦ avols ⊆ avols1
Rule argument
Additional information