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Proof step 4 from proof subset (previous step)

b ∈ s ⊦ b = a, a ∈ s, s ⊆ s ++ a ++ b

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2)

Rule argument

spec: <toplevel> name: Subset seq: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2) subst: [s2, s1] → [s ++ a ++ b, s] flags: <all paths>

Additional information

Used simplifier rules: In-insert: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; Used terms: s ++ a ++ b, s