YES

Problem 1: 

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

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 A(a(x1:S)) -> x1:S
 A(b(x1:S)) -> b(a(B(A(x1:S))))
 B(a(x1:S)) -> a(b(A(B(x1:S))))
 B(b(x1:S)) -> 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)
 A#(b(x1:S)) -> B#(A(x1:S))
 B#(a(x1:S)) -> A#(B(x1:S))
 B#(a(x1:S)) -> B#(x1:S)
-> Rules:
 A(a(x1:S)) -> x1:S
 A(b(x1:S)) -> b(a(B(A(x1:S))))
 B(a(x1:S)) -> a(b(A(B(x1:S))))
 B(b(x1:S)) -> x1:S

Problem 1: 

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

Problem 1: 

Narrowing Processor:
-> Pairs:
 A#(b(x1:S)) -> A#(x1:S)
 A#(b(x1:S)) -> B#(A(x1:S))
 B#(a(x1:S)) -> A#(B(x1:S))
 B#(a(x1:S)) -> B#(x1:S)
-> Rules:
 A(a(x1:S)) -> x1:S
 A(b(x1:S)) -> b(a(B(A(x1:S))))
 B(a(x1:S)) -> a(b(A(B(x1:S))))
 B(b(x1:S)) -> x1:S
->Narrowed Pairs:
->->Original Pair:
 A#(b(x1:S)) -> B#(A(x1:S))
->-> Narrowed pairs:
 A#(b(a(x1:S))) -> B#(x1:S)
 A#(b(b(x1:S))) -> B#(b(a(B(A(x1:S)))))
->->Original Pair:
 B#(a(x1:S)) -> A#(B(x1:S))
->-> Narrowed pairs:
 B#(a(a(x1:S))) -> A#(a(b(A(B(x1:S)))))
 B#(a(b(x1:S))) -> A#(x1:S)

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.