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Proof step 7 from proof first-01 (previous step)

⊦ (∀ a0. ¬ a0 ∈ y ∨ ¬ a0 ∈ x) ↔ (∀ a. ¬(a ∈ y ∧ a ∈ x)), a ∈ x

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a0 ∈ y, a0 ∈ x, ∀ a0. ¬ a0 ∈ y ∨ ¬ a0 ∈ x ⊦ a ∈ x

Rule argument

Additional information