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Proof step 8 from proof diff-dups (previous step)

dups((a ' + x0) \ y), ∀ y0. ¬ dups(x0 \ y0) ⊦ dups(x0), a ∈ x0, a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 9:
dups(x0 \ y), ∀ y0. ¬ dups(x0 \ y0) ⊦ dups(x0), a ∈ y, a ∈ x0

Rule argument

Additional information

Used simplifier rules: diff-n: ⊦ ¬ a ∈ y → (a ' + x) \ y = a + (x \ y);%diff-in: ⊦ a ∈ x \ y ↔ a ∈ x ∧ ¬ a ∈ y;% ⊦ dups(a ' + x) ↔ a ∈ x ∨ dups(x); (s)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)