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Proof step 1 from proof subset-app-04 (conclusion)

⊦ x + y ⊆ z + y ↔ x ⊆ z + 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 + y → a ∈ z + y) ↔ x ⊆ z + y

Rule argument

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

Additional information

Used terms: z + y, x + y