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Proof step 5 from proof notin-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a0 ∈ y, a ∈ y, a ∈ x0 \ y, ∀ a1, y0. ¬ a1 ∈ x0 \ y0 ∨ ¬ a1 ∈ y0 ⊦

Rule argument

Additional information

Used simplifier rules: diff-y: ⊦ a ∈ y → (a ' + x) \ y = x \ y;%a: ⊦ (x + y) + z = x + y + z; from specification list