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Proof step 4 from proof subset-app-04 (previous step)

⊦ (∀ a. a ∈ x → a ∈ z ∨ a ∈ y) ↔ x ⊆ z + 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:
⊦ (∀ a. a ∈ x → a ∈ z ∨ a ∈ y) ↔ (∀ a. a ∈ x → a ∈ z + y)

Rule argument

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

Additional information

Used terms: z + y, x