YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S)
(RULES
if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S))
if(ttrue,x:S,y:S) -> 0
le(0,y:S) -> ttrue
le(s(x:S),0) -> ffalse
le(s(x:S),s(y:S)) -> le(x:S,y:S)
minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S)
p(0) -> 0
p(s(x:S)) -> x:S
) 
(STRATEGY INNERMOST)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 IF(ffalse,x:S,y:S) -> MINUS(p(x:S),y:S)
 IF(ffalse,x:S,y:S) -> P(x:S)
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
 MINUS(x:S,y:S) -> IF(le(x:S,y:S),x:S,y:S)
 MINUS(x:S,y:S) -> LE(x:S,y:S)
-> Rules:
 if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S))
 if(ttrue,x:S,y:S) -> 0
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S)
 p(0) -> 0
 p(s(x:S)) -> x:S

Problem 1: 

SCC Processor:
-> Pairs:
 IF(ffalse,x:S,y:S) -> MINUS(p(x:S),y:S)
 IF(ffalse,x:S,y:S) -> P(x:S)
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
 MINUS(x:S,y:S) -> IF(le(x:S,y:S),x:S,y:S)
 MINUS(x:S,y:S) -> LE(x:S,y:S)
-> Rules:
 if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S))
 if(ttrue,x:S,y:S) -> 0
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
->->-> Rules:
 if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S))
 if(ttrue,x:S,y:S) -> 0
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S)
 p(0) -> 0
 p(s(x:S)) -> x:S
->->Cycle:
->->-> Pairs:
 IF(ffalse,x:S,y:S) -> MINUS(p(x:S),y:S)
 MINUS(x:S,y:S) -> IF(le(x:S,y:S),x:S,y:S)
->->-> Rules:
 if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S))
 if(ttrue,x:S,y:S) -> 0
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S)
 p(0) -> 0
 p(s(x:S)) -> x:S


The problem is decomposed in 2 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 LE(s(x:S),s(y:S)) -> LE(x:S,y:S)
-> Rules:
 if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S))
 if(ttrue,x:S,y:S) -> 0
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Projection:
 pi(LE) = 1

Problem 1.1: 

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

The problem is finite.

Problem 1.2: 

Reduction Pairs Processor:
-> Pairs:
 IF(ffalse,x:S,y:S) -> MINUS(p(x:S),y:S)
 MINUS(x:S,y:S) -> IF(le(x:S,y:S),x:S,y:S)
-> Rules:
 if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S))
 if(ttrue,x:S,y:S) -> 0
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S)
 p(0) -> 0
 p(s(x:S)) -> x:S
-> Usable rules:
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[if](X1,X2,X3) = 0
[le](X1,X2) = 1/2.X1 + 1/2
[minus](X1,X2) = 0
[p](X) = 1/2.X
[0] = 0
[fSNonEmpty] = 0
[false] = 1
[s](X) = 2.X + 2
[true] = 0
[IF](X1,X2,X3) = 2.X1 + X2 + 2.X3
[LE](X1,X2) = 0
[MINUS](X1,X2) = 2.X1 + 2.X2 + 1
[P](X) = 0

Problem 1.2: 

SCC Processor:
-> Pairs:
 MINUS(x:S,y:S) -> IF(le(x:S,y:S),x:S,y:S)
-> Rules:
 if(ffalse,x:S,y:S) -> s(minus(p(x:S),y:S))
 if(ttrue,x:S,y:S) -> 0
 le(0,y:S) -> ttrue
 le(s(x:S),0) -> ffalse
 le(s(x:S),s(y:S)) -> le(x:S,y:S)
 minus(x:S,y:S) -> if(le(x:S,y:S),x:S,y:S)
 p(0) -> 0
 p(s(x:S)) -> x:S
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.