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Proof step 1 from proof sub-ntrans (conclusion)

⊦ s ⊆ s0 ∧ ¬ s ⊆ s1 → ¬ s0 ⊆ s1

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: sub-trans: ⊦ s ⊆ s0 ∧ s0 ⊆ s1 → s ⊆ s1;%sub-ntrans-01: ⊦ s0 ⊆ s1 ∧ ¬ s ⊆ s1 → ¬ s ⊆ s0;