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Proof step 1 from proof diff-empty-sub (conclusion)

⊦ x \ y = [] ↔ x ⊆ y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ x \ y = [] ↔ (∀ a. a ∈ x → a ∈ y)
Proof step 3:
⊦ (∀ a. a ∈ x → a ∈ y) ↔ x ⊆ y

Rule argument

spec: list-set inst: name: diff-empty-all seq: ⊦ x \ y = [] ↔ (∀ a. a ∈ x → a ∈ y) subst: [y, x] → [y, x] flags: rewrite, current

Additional information

Used terms: y, x