YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S)
(RULES
-(0,y:S) -> 0
-(x:S,0) -> x:S
-(x:S,s(y:S)) -> if(greater(x:S,s(y:S)),s(-(x:S,p(s(y:S)))),0)
p(0) -> 0
p(s(x:S)) -> x:S
)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 -#(x:S,s(y:S)) -> -#(x:S,p(s(y:S)))
 -#(x:S,s(y:S)) -> P(s(y:S))
-> Rules:
 -(0,y:S) -> 0
 -(x:S,0) -> x:S
 -(x:S,s(y:S)) -> if(greater(x:S,s(y:S)),s(-(x:S,p(s(y:S)))),0)
 p(0) -> 0
 p(s(x:S)) -> x:S

Problem 1: 

SCC Processor:
-> Pairs:
 -#(x:S,s(y:S)) -> -#(x:S,p(s(y:S)))
 -#(x:S,s(y:S)) -> P(s(y:S))
-> Rules:
 -(0,y:S) -> 0
 -(x:S,0) -> x:S
 -(x:S,s(y:S)) -> if(greater(x:S,s(y:S)),s(-(x:S,p(s(y:S)))),0)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 -#(x:S,s(y:S)) -> -#(x:S,p(s(y:S)))
->->-> Rules:
 -(0,y:S) -> 0
 -(x:S,0) -> x:S
 -(x:S,s(y:S)) -> if(greater(x:S,s(y:S)),s(-(x:S,p(s(y:S)))),0)
 p(0) -> 0
 p(s(x:S)) -> x:S

Problem 1: 

Reduction Pair Processor:
-> Pairs:
 -#(x:S,s(y:S)) -> -#(x:S,p(s(y:S)))
-> Rules:
 -(0,y:S) -> 0
 -(x:S,0) -> x:S
 -(x:S,s(y:S)) -> if(greater(x:S,s(y:S)),s(-(x:S,p(s(y:S)))),0)
 p(0) -> 0
 p(s(x:S)) -> x:S
-> Usable rules:
 p(0) -> 0
 p(s(x:S)) -> x:S
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[p](X) = 1/2.X + 1/2
[0] = 1/2
[s](X) = 2.X + 2
[-#](X1,X2) = 2.X2

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 -(0,y:S) -> 0
 -(x:S,0) -> x:S
 -(x:S,s(y:S)) -> if(greater(x:S,s(y:S)),s(-(x:S,p(s(y:S)))),0)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.