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Proof step 1 from proof sub-add (conclusion)

⊦ a ∈ s0 → (s ++ a ⊆ s0 ↔ s ⊆ s0)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: Subset: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2);%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);