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Proof step 4 from proof diff-dups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
dups((a ' + x0) \ y), ∀ y0. ¬ dups(x0 \ y0) ⊦ dups(x0), a ∈ x0

Rule argument

Additional information

Used simplifier rules: a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)% ⊦ a + x = a ' + x; (s)