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Proof step 3 from proof notin-01 (previous step)

∀ a, y. ¬(a ∈ x0 \ y ∧ a ∈ y), a ∈ (a0 + x0) \ y, a ∈ y ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
a ∈ y, a ∈ (a0 ' + x0) \ y, ∀ a1, y0. ¬ a1 ∈ x0 \ y0 ∨ ¬ a1 ∈ y0 ⊦

Rule argument

Additional information

Used simplifier rules: c: ⊦ a + x = a ' + x; from specification list