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Proof step 3 from proof f (previous step)

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

Heuristic cut applied rule cut formula

And obtained the following new goals

Proof step 4:
∀ a. dups(x0 + a ') ↔ a ∈ x0 ∨ dups(x0) ⊦ dups(x0), a0 = a ∨ dups(x0 + a ') ↔ a = a0 ∨ a ∈ x0 ∨ dups(x0), a0 ∈ x0

Rule argument

fma: dups(x0)

Additional information

Used simplifier rules: app-05: ⊦ dups(x) → dups(x + y);%a: ⊦ (x + y) + z = x + y + z; from specification list Used formulas: dups(x0)