YES

Problem 1: 

(VAR v_NonEmpty:S x:S y:S)
(RULES
-(s(x:S),s(y:S)) -> -(x:S,y:S)
-(x:S,0) -> x:S
if(ffalse,x:S,y:S) -> y:S
if(ttrue,x:S,y:S) -> x:S
leq(0,y:S) -> ttrue
leq(s(x:S),0) -> ffalse
leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
mod(0,y:S) -> 0
mod(s(x:S),0) -> 0
mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 -#(s(x:S),s(y:S)) -> -#(x:S,y:S)
 LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S)
 MOD(s(x:S),s(y:S)) -> -#(s(x:S),s(y:S))
 MOD(s(x:S),s(y:S)) -> IF(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
 MOD(s(x:S),s(y:S)) -> LEQ(y:S,x:S)
 MOD(s(x:S),s(y:S)) -> MOD(-(s(x:S),s(y:S)),s(y:S))
-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))

Problem 1: 

SCC Processor:
-> Pairs:
 -#(s(x:S),s(y:S)) -> -#(x:S,y:S)
 LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S)
 MOD(s(x:S),s(y:S)) -> -#(s(x:S),s(y:S))
 MOD(s(x:S),s(y:S)) -> IF(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
 MOD(s(x:S),s(y:S)) -> LEQ(y:S,x:S)
 MOD(s(x:S),s(y:S)) -> MOD(-(s(x:S),s(y:S)),s(y:S))
-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S)
->->-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
->->Cycle:
->->-> Pairs:
 -#(s(x:S),s(y:S)) -> -#(x:S,y:S)
->->-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
->->Cycle:
->->-> Pairs:
 MOD(s(x:S),s(y:S)) -> MOD(-(s(x:S),s(y:S)),s(y:S))
->->-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))


The problem is decomposed in 3 subproblems.

Problem 1.1: 

Subterm Processor:
-> Pairs:
 LEQ(s(x:S),s(y:S)) -> LEQ(x:S,y:S)
-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
->Projection:
 pi(LEQ) = 1

Problem 1.1: 

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

The problem is finite.

Problem 1.2: 

Subterm Processor:
-> Pairs:
 -#(s(x:S),s(y:S)) -> -#(x:S,y:S)
-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
->Projection:
 pi(-#) = 1

Problem 1.2: 

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

The problem is finite.

Problem 1.3: 

Narrowing Processor:
-> Pairs:
 MOD(s(x:S),s(y:S)) -> MOD(-(s(x:S),s(y:S)),s(y:S))
-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
->Narrowed Pairs:
->->Original Pair:
 MOD(s(x:S),s(y:S)) -> MOD(-(s(x:S),s(y:S)),s(y:S))
->-> Narrowed pairs:
 MOD(s(x:S),s(y:S)) -> MOD(-(x:S,y:S),s(y:S))

Problem 1.3: 

SCC Processor:
-> Pairs:
 MOD(s(x:S),s(y:S)) -> MOD(-(x:S,y:S),s(y:S))
-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 MOD(s(x:S),s(y:S)) -> MOD(-(x:S,y:S),s(y:S))
->->-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))

Problem 1.3: 

Reduction Pairs Processor:
-> Pairs:
 MOD(s(x:S),s(y:S)) -> MOD(-(x:S,y:S),s(y:S))
-> Rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
 if(ffalse,x:S,y:S) -> y:S
 if(ttrue,x:S,y:S) -> x:S
 leq(0,y:S) -> ttrue
 leq(s(x:S),0) -> ffalse
 leq(s(x:S),s(y:S)) -> leq(x:S,y:S)
 mod(0,y:S) -> 0
 mod(s(x:S),0) -> 0
 mod(s(x:S),s(y:S)) -> if(leq(y:S,x:S),mod(-(s(x:S),s(y:S)),s(y:S)),s(x:S))
-> Usable rules:
 -(s(x:S),s(y:S)) -> -(x:S,y:S)
 -(x:S,0) -> x:S
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[-](X1,X2) = 2.X1 + 1
[if](X1,X2,X3) = 0
[leq](X1,X2) = 0
[mod](X1,X2) = 0
[0] = 0
[fSNonEmpty] = 0
[false] = 0
[s](X) = 2.X + 2
[true] = 0
[-#](X1,X2) = 0
[IF](X1,X2,X3) = 0
[LEQ](X1,X2) = 0
[MOD](X1,X2) = 2.X1

Problem 1.3: 

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

The problem is finite.