YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S)
(RULES
int(0,0) -> .(0,nil)
int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
int(s(x:S),0) -> nil
int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
int_list(nil) -> nil
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 int(0,0) -> .(0,nil)
 int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
 int(s(x:S),0) -> nil
 int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
 int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
 int_list(nil) -> nil
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 INT(0,s(y:S)) -> INT(s(0),s(y:S))
 INT(s(x:S),s(y:S)) -> INT(x:S,y:S)
 INT(s(x:S),s(y:S)) -> INT_LIST(int(x:S,y:S))
 INT_LIST(.(x:S,y:S)) -> INT_LIST(y:S)
-> Rules:
 int(0,0) -> .(0,nil)
 int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
 int(s(x:S),0) -> nil
 int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
 int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
 int_list(nil) -> nil

Problem 1: 

SCC Processor:
-> Pairs:
 INT(0,s(y:S)) -> INT(s(0),s(y:S))
 INT(s(x:S),s(y:S)) -> INT(x:S,y:S)
 INT(s(x:S),s(y:S)) -> INT_LIST(int(x:S,y:S))
 INT_LIST(.(x:S,y:S)) -> INT_LIST(y:S)
-> Rules:
 int(0,0) -> .(0,nil)
 int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
 int(s(x:S),0) -> nil
 int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
 int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
 int_list(nil) -> nil
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 INT_LIST(.(x:S,y:S)) -> INT_LIST(y:S)
->->-> Rules:
 int(0,0) -> .(0,nil)
 int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
 int(s(x:S),0) -> nil
 int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
 int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
 int_list(nil) -> nil
->->Cycle:
->->-> Pairs:
 INT(0,s(y:S)) -> INT(s(0),s(y:S))
 INT(s(x:S),s(y:S)) -> INT(x:S,y:S)
->->-> Rules:
 int(0,0) -> .(0,nil)
 int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
 int(s(x:S),0) -> nil
 int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
 int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
 int_list(nil) -> nil


The problem is decomposed in 2 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 INT_LIST(.(x:S,y:S)) -> INT_LIST(y:S)
-> Rules:
 int(0,0) -> .(0,nil)
 int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
 int(s(x:S),0) -> nil
 int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
 int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
 int_list(nil) -> nil
->Projection:
 pi(INT_LIST) = 1

Problem 1.1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 int(0,0) -> .(0,nil)
 int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
 int(s(x:S),0) -> nil
 int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
 int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
 int_list(nil) -> nil
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

Subterm Processor:
-> Pairs:
 INT(0,s(y:S)) -> INT(s(0),s(y:S))
 INT(s(x:S),s(y:S)) -> INT(x:S,y:S)
-> Rules:
 int(0,0) -> .(0,nil)
 int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
 int(s(x:S),0) -> nil
 int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
 int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
 int_list(nil) -> nil
->Projection:
 pi(INT) = 2

Problem 1.2: 

SCC Processor:
-> Pairs:
 INT(0,s(y:S)) -> INT(s(0),s(y:S))
-> Rules:
 int(0,0) -> .(0,nil)
 int(0,s(y:S)) -> .(0,int(s(0),s(y:S)))
 int(s(x:S),0) -> nil
 int(s(x:S),s(y:S)) -> int_list(int(x:S,y:S))
 int_list(.(x:S,y:S)) -> .(s(x:S),int_list(y:S))
 int_list(nil) -> nil
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.