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Proof step 15 from proof diff-unchanged (previous step)

x0 \ y = x0 ↔ (∀ a. ¬ a ∈ x0 ∨ ¬ a ∈ y), ∀ a0. ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y ⊦ x0 \ y = x0, a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
a0 ∈ x0, a0 ∈ y, ∀ a0. ¬ a0 ∈ x0 ∨ ¬ a0 ∈ y ⊦ x0 \ y = x0, a ∈ y

Rule argument

Additional information