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Proof step 4 from proof subset-rec (previous step)

⊦ (∀ a0. a0 = a ∨ a0 ∈ x → a0 ∈ y) ↔ a ∈ y ∧ x ⊆ y

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ x ⊆ y ↔ (∀ a. a ∈ x → a ∈ y)
Proof step 6:
⊦ (∀ a0. a0 = a ∨ a0 ∈ x → a0 ∈ y) ↔ a ∈ y ∧ (∀ a. a ∈ x → a ∈ y)

Rule argument

spec: list-set inst: name: subset seq: ⊦ x ⊆ y ↔ (∀ a. a ∈ x → a ∈ y) subst: [y, x] → [y, x] flags: rewrite, current

Additional information

Used terms: y, x