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Proof step 3 from proof subset-app-04 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (∀ a. a ∈ x → a ∈ z ∨ a ∈ y) ↔ x ⊆ z + y

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%in-append: ⊦ a ∈ x + y ↔ ¬ ¬(a ∈ x ∨ a ∈ y); from specification list