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Proof step 3 from proof subset-app-03 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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