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Proof step 4 from proof f (previous step)

∀ a. dups(x0 + a ') ↔ a ∈ x0 ∨ dups(x0) ⊦ dups(x0), a0 = a ∨ dups(x0 + a ') ↔ a = a0 ∨ a ∈ x0 ∨ dups(x0), a0 ∈ x0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
dups(x0 + a '), ∀ a1. dups(x0 + a1 ') → a1 ∈ x0 ⊦ a = a0, dups(x0), a0 ∈ x0, a ∈ x0

Rule argument

Additional information

Used simplifier rules: l: ⊦ a ∈ x → dups(x + a ');%a: ⊦ (x + y) + z = x + y + z; from specification list