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Proof step 6 from proof diff-dups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
dups(x0 \ y), a ∈ y, ∀ y0. ¬ dups(x0 \ y0) ⊦ dups(x0), a ∈ x0

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