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Proof step 4 from proof subset-append-02 (previous step)

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

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 ∈ y → a ∈ z) ↔ (∀ a. a ∈ x → a ∈ z) ∧ y ⊆ z

Rule argument

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

Additional information

Used terms: z, x