YES

Problem 1: 

(VAR v_NonEmpty:S X:S X1:S X2:S X3:S)
(RULES
a__b -> a
a__b -> b
a__f(a,X:S,X:S) -> a__f(X:S,a__b,b)
a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
mark(a) -> a
mark(b) -> a__b
mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
) 
(STRATEGY INNERMOST)

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 A__F(a,X:S,X:S) -> A__B
 A__F(a,X:S,X:S) -> A__F(X:S,a__b,b)
 MARK(b) -> A__B
 MARK(f(X1:S,X2:S,X3:S)) -> A__F(X1:S,mark(X2:S),X3:S)
 MARK(f(X1:S,X2:S,X3:S)) -> MARK(X2:S)
-> Rules:
 a__b -> a
 a__b -> b
 a__f(a,X:S,X:S) -> a__f(X:S,a__b,b)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(a) -> a
 mark(b) -> a__b
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)

Problem 1: 

SCC Processor:
-> Pairs:
 A__F(a,X:S,X:S) -> A__B
 A__F(a,X:S,X:S) -> A__F(X:S,a__b,b)
 MARK(b) -> A__B
 MARK(f(X1:S,X2:S,X3:S)) -> A__F(X1:S,mark(X2:S),X3:S)
 MARK(f(X1:S,X2:S,X3:S)) -> MARK(X2:S)
-> Rules:
 a__b -> a
 a__b -> b
 a__f(a,X:S,X:S) -> a__f(X:S,a__b,b)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(a) -> a
 mark(b) -> a__b
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 A__F(a,X:S,X:S) -> A__F(X:S,a__b,b)
->->-> Rules:
 a__b -> a
 a__b -> b
 a__f(a,X:S,X:S) -> a__f(X:S,a__b,b)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(a) -> a
 mark(b) -> a__b
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
->->Cycle:
->->-> Pairs:
 MARK(f(X1:S,X2:S,X3:S)) -> MARK(X2:S)
->->-> Rules:
 a__b -> a
 a__b -> b
 a__f(a,X:S,X:S) -> a__f(X:S,a__b,b)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(a) -> a
 mark(b) -> a__b
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)


The problem is decomposed in 2 subproblems.

Problem 1.1: 

Reduction Pairs Processor:
-> Pairs:
 A__F(a,X:S,X:S) -> A__F(X:S,a__b,b)
-> Rules:
 a__b -> a
 a__b -> b
 a__f(a,X:S,X:S) -> a__f(X:S,a__b,b)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(a) -> a
 mark(b) -> a__b
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
-> Usable rules:
 a__b -> a
 a__b -> b
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[a__b] = 1
[a__f](X1,X2,X3) = 0
[mark](X) = 0
[a] = 1
[b] = 0
[f](X1,X2,X3) = 0
[fSNonEmpty] = 0
[A__B] = 0
[A__F](X1,X2,X3) = 2.X1 + X2 + 2.X3
[MARK](X) = 0

Problem 1.1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 a__b -> a
 a__b -> b
 a__f(a,X:S,X:S) -> a__f(X:S,a__b,b)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(a) -> a
 mark(b) -> a__b
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.

Problem 1.2: 

Subterm Processor:
-> Pairs:
 MARK(f(X1:S,X2:S,X3:S)) -> MARK(X2:S)
-> Rules:
 a__b -> a
 a__b -> b
 a__f(a,X:S,X:S) -> a__f(X:S,a__b,b)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(a) -> a
 mark(b) -> a__b
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
->Projection:
 pi(MARK) = 1

Problem 1.2: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 a__b -> a
 a__b -> b
 a__f(a,X:S,X:S) -> a__f(X:S,a__b,b)
 a__f(X1:S,X2:S,X3:S) -> f(X1:S,X2:S,X3:S)
 mark(a) -> a
 mark(b) -> a__b
 mark(f(X1:S,X2:S,X3:S)) -> a__f(X1:S,mark(X2:S),X3:S)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.