YES

Problem 1: 

(VAR v_NonEmpty:S x1:S)
(RULES
0(0(x1:S)) -> 0(x1:S)
0(1(x1:S)) -> 1(x1:S)
1(0(x1:S)) -> 0(0(0(1(x1:S))))
1(1(x1:S)) -> 0(0(0(0(x1:S))))
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 1#(0(x1:S)) -> 0#(0(0(1(x1:S))))
 1#(0(x1:S)) -> 0#(0(1(x1:S)))
 1#(0(x1:S)) -> 0#(1(x1:S))
 1#(0(x1:S)) -> 1#(x1:S)
 1#(1(x1:S)) -> 0#(0(0(0(x1:S))))
 1#(1(x1:S)) -> 0#(0(0(x1:S)))
 1#(1(x1:S)) -> 0#(0(x1:S))
 1#(1(x1:S)) -> 0#(x1:S)
-> Rules:
 0(0(x1:S)) -> 0(x1:S)
 0(1(x1:S)) -> 1(x1:S)
 1(0(x1:S)) -> 0(0(0(1(x1:S))))
 1(1(x1:S)) -> 0(0(0(0(x1:S))))

Problem 1: 

SCC Processor:
-> Pairs:
 1#(0(x1:S)) -> 0#(0(0(1(x1:S))))
 1#(0(x1:S)) -> 0#(0(1(x1:S)))
 1#(0(x1:S)) -> 0#(1(x1:S))
 1#(0(x1:S)) -> 1#(x1:S)
 1#(1(x1:S)) -> 0#(0(0(0(x1:S))))
 1#(1(x1:S)) -> 0#(0(0(x1:S)))
 1#(1(x1:S)) -> 0#(0(x1:S))
 1#(1(x1:S)) -> 0#(x1:S)
-> Rules:
 0(0(x1:S)) -> 0(x1:S)
 0(1(x1:S)) -> 1(x1:S)
 1(0(x1:S)) -> 0(0(0(1(x1:S))))
 1(1(x1:S)) -> 0(0(0(0(x1:S))))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 1#(0(x1:S)) -> 1#(x1:S)
->->-> Rules:
 0(0(x1:S)) -> 0(x1:S)
 0(1(x1:S)) -> 1(x1:S)
 1(0(x1:S)) -> 0(0(0(1(x1:S))))
 1(1(x1:S)) -> 0(0(0(0(x1:S))))

Problem 1: 

Subterm Processor:
-> Pairs:
 1#(0(x1:S)) -> 1#(x1:S)
-> Rules:
 0(0(x1:S)) -> 0(x1:S)
 0(1(x1:S)) -> 1(x1:S)
 1(0(x1:S)) -> 0(0(0(1(x1:S))))
 1(1(x1:S)) -> 0(0(0(0(x1:S))))
->Projection:
 pi(1#) = 1

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 0(0(x1:S)) -> 0(x1:S)
 0(1(x1:S)) -> 1(x1:S)
 1(0(x1:S)) -> 0(0(0(1(x1:S))))
 1(1(x1:S)) -> 0(0(0(0(x1:S))))
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.