YES

Problem 1: 

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

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 a(b(x:S)) -> b(a(x:S))
 a(c(x:S)) -> x: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(x:S)) -> A(x:S)
-> Rules:
 a(b(x:S)) -> b(a(x:S))
 a(c(x:S)) -> x:S

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.