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Proof step 2 from proof dups-app (previous step)

∀ y. dups(x0 + y) ↔ dups(x0) ∨ dups(y) ∨ filter(x0, y) ≠ [] ⊦ dups((a + x0) + y) ↔ dups(a + x0) ∨ dups(y) ∨ filter(a + x0, y) ≠ []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
∀ y. dups(x0 + y) ↔ dups(x0) ∨ dups(y) ∨ filter(x0, y) ≠ [] ⊦ a ∈ y ∨ dups(x0 + y) ↔ dups(x0) ∨ dups(y) ∨ filter(a ' + x0, y) ≠ [], a ∈ x0

Rule argument

Additional information

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