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Proof step 9 from proof suband (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a ∈ x ∨ a ∈ y, ∀ a. a ∈ x → a ∈ z, ∀ a. a ∈ y → a ∈ z ⊦ a ∈ z

Rule argument

Additional information