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Proof step 11 from proof subset-rec (previous step)

a ∈ y ∧ (∀ a. a ∈ x → a ∈ y) ⊦ ∀ a0. a0 = a ∨ a0 ∈ x → a0 ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
a ∈ y, a0 ∈ x, ∀ a. a ∈ x → a ∈ y ⊦ a0 ∈ y

Rule argument

Additional information