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Proof step 6 from proof subset-insert (previous step)

c ∈ s ⊦ b = a, a ∈ s, s ⊆ s ++ a ++ b ++ c, b ∈ s

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2)

Rule argument

spec: inst: name: Subset seq: ⊦ s1 ⊆ s2 ↔ (∀ a. a ∈ s1 → a ∈ s2) subst: [s2, s1] → [s ++ a ++ b ++ c, s] flags: rewrite, current

Additional information

Used simplifier rules: add: ⊦ ¬ a ∈ s ++ b ↔ a ≠ b ∧ ¬ a ∈ s;%rax: ⊦ a ∈ s ++ b ↔ ¬ ¬(a = b ∨ a ∈ s);%in: ⊦ a ∈ s → a ∈ s ++ b; Used terms: s ++ a ++ b ++ c, s