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Proof step 3 from proof diff-dups (previous step)

∀ y. dups(x0 \ y) → dups(x0), dups((a + x0) \ y) ⊦ dups(a + x0)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%c: ⊦ a + x = a ' + x; from specification list