YES

Problem 1: 

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

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 a(b(x1:S)) -> b(b(a(x1:S)))
 c(b(x1:S)) -> b(c(c(x1:S)))
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

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

Problem 1: 

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


The problem is decomposed in 2 subproblems.

Problem 1.1: 

Reduction Pairs Processor:
-> Pairs:
 C(b(x1:S)) -> C(c(x1:S))
 C(b(x1:S)) -> C(x1:S)
-> Rules:
 a(b(x1:S)) -> b(b(a(x1:S)))
 c(b(x1:S)) -> b(c(c(x1:S)))
-> Usable rules:
 c(b(x1:S)) -> b(c(c(x1:S)))
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[a](X) = 0
[c](X) = X
[b](X) = X + 2
[fSNonEmpty] = 0
[A](X) = 0
[C](X) = 2.X

Problem 1.1: 

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

Problem 1.1: 

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

Problem 1.1: 

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

The problem is finite.

Problem 1.2: 

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

Problem 1.2: 

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

The problem is finite.