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Proof step 1 from proof sub-del-01 (conclusion)

⊦ s ⊆ s0 ∧ ¬ a ∈ s → s ⊆ s0 -- a

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: subdef: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2);%sub-del-04: ⊦ s ⊆ s0 -- a ↔ s ⊆ s0 ∧ ¬ a ∈ s;