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Proof step 1 from proof sub-ntrans-01 (conclusion)
⊦ s0 ⊆ s1 ∧ ¬ s ⊆ s1 → ¬ s ⊆ s0
System applied rule normalize
And obtained the following new goals
Proof step 2
:
s ⊆ s0, s0 ⊆ s1 ⊦ s ⊆ s1
Rule argument
Additional information