YES

Problem 1: 

(VAR v_NonEmpty:S x1:S)
(RULES
a(x1:S) -> b(b(x1:S))
b(b(b(x1:S))) -> a(x1:S)
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 A(x1:S) -> B(b(x1:S))
 A(x1:S) -> B(x1:S)
 B(b(b(x1:S))) -> A(x1:S)
-> Rules:
 a(x1:S) -> b(b(x1:S))
 b(b(b(x1:S))) -> a(x1:S)

Problem 1: 

SCC Processor:
-> Pairs:
 A(x1:S) -> B(b(x1:S))
 A(x1:S) -> B(x1:S)
 B(b(b(x1:S))) -> A(x1:S)
-> Rules:
 a(x1:S) -> b(b(x1:S))
 b(b(b(x1:S))) -> a(x1:S)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 A(x1:S) -> B(b(x1:S))
 A(x1:S) -> B(x1:S)
 B(b(b(x1:S))) -> A(x1:S)
->->-> Rules:
 a(x1:S) -> b(b(x1:S))
 b(b(b(x1:S))) -> a(x1:S)

Problem 1: 

Reduction Pair Processor:
-> Pairs:
 A(x1:S) -> B(b(x1:S))
 A(x1:S) -> B(x1:S)
 B(b(b(x1:S))) -> A(x1:S)
-> Rules:
 a(x1:S) -> b(b(x1:S))
 b(b(b(x1:S))) -> a(x1:S)
-> Usable rules:
 a(x1:S) -> b(b(x1:S))
 b(b(b(x1:S))) -> a(x1:S)
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[a](X) = X + 2
[b](X) = X + 1
[A](X) = X + 2
[B](X) = X

Problem 1: 

SCC Processor:
-> Pairs:
 A(x1:S) -> B(x1:S)
 B(b(b(x1:S))) -> A(x1:S)
-> Rules:
 a(x1:S) -> b(b(x1:S))
 b(b(b(x1:S))) -> a(x1:S)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 A(x1:S) -> B(x1:S)
 B(b(b(x1:S))) -> A(x1:S)
->->-> Rules:
 a(x1:S) -> b(b(x1:S))
 b(b(b(x1:S))) -> a(x1:S)

Problem 1: 

Subterm Processor:
-> Pairs:
 A(x1:S) -> B(x1:S)
 B(b(b(x1:S))) -> A(x1:S)
-> Rules:
 a(x1:S) -> b(b(x1:S))
 b(b(b(x1:S))) -> a(x1:S)
->Projection:
 pi(A) = 1
 pi(B) = 1

Problem 1: 

SCC Processor:
-> Pairs:
 A(x1:S) -> B(x1:S)
-> Rules:
 a(x1:S) -> b(b(x1:S))
 b(b(b(x1:S))) -> a(x1:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.