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Proof step 1 from proof subset-ins (conclusion)

⊦ s ++ a ⊆ s0 ++ a ++ b ↔ s ⊆ s0 ++ a ++ b

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: sub-add-02: ⊦ s ++ a ⊆ s0 → s ⊆ s0;%sub-add-03: ⊦ s ++ a ⊆ s0 ↔ s ⊆ s0 ∧ a ∈ s0;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);