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Proof step 1 from proof diff-subseteq-subseteq (conclusion)
⊦ as1 ⊆ as2 → (as0 \ as1 ⊆ as2 ↔ as0 ⊆ as2)
System applied rule normalize
And obtained the following new goals
Proof step 2
:
as1 ⊆ as2 ⊦ as0 \ as1 ⊆ as2 ↔ as0 ⊆ as2
Rule argument
Additional information