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Proof step 1 from proof sub-notin (conclusion)

⊦ x ⊆ y ∧ ¬ a ∈ y → ¬ a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
a ∈ x, ∀ a. a ∈ x → a ∈ y ⊦ a ∈ y

Rule argument

Additional information

Used simplifier rules: subset: ⊦ x ⊆ y ↔ (∀ a. a ∈ x → a ∈ y);