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

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
∀ y. dups(x0 + y) ↔ dups(x0) ∨ dups(y) ∨ filter(x0, y) ≠ [] ⊦ dups(x0 + y) ↔ dups(x0) ∨ dups(y) ∨ filter(x0, y) ≠ [], a ∈ y, a ∈ x0

Rule argument

Additional information

Used simplifier rules: filt-n: ⊦ ¬ a ∈ y → filter(a ' + x, y) = filter(x, y);%a: ⊦ (x + y) + z = x + y + z; from specification list