YES

Problem 1: 

(VAR x y)
(THEORY 
(AC plus))
(RULES
L(T(x)) -> L(x)
L(plus(T(y),x)) -> plus(L(plus(x,y)),L(y))
T(T(x)) -> T(x)
T(plus(T(y),x)) -> plus(T(plus(x,y)),T(y))
plus(T(plus(x,y)),x) -> T(plus(x,y))
plus(T(x),x) -> T(x)
plus(x,0) -> x
plus(x,x) -> x
)

Problem 1: 

Reduction Order Processor:
-> Rules:
 L(T(x)) -> L(x)
 L(plus(T(y),x)) -> plus(L(plus(x,y)),L(y))
 T(T(x)) -> T(x)
 T(plus(T(y),x)) -> plus(T(plus(x,y)),T(y))
 plus(T(plus(x,y)),x) -> T(plus(x,y))
 plus(T(x),x) -> T(x)
 plus(x,0) -> x
 plus(x,x) -> x
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[L](X) = X + 1
[T](X) = 2.X + 2
[plus](X1,X2) = X1 + X2 + 1
[0] = 0

Problem 1: 

Reduction Order Processor:
-> Rules:
 L(plus(T(y),x)) -> plus(L(plus(x,y)),L(y))
 T(T(x)) -> T(x)
 T(plus(T(y),x)) -> plus(T(plus(x,y)),T(y))
 plus(T(plus(x,y)),x) -> T(plus(x,y))
 plus(T(x),x) -> T(x)
 plus(x,0) -> x
 plus(x,x) -> x
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[L](X) = 2.X + 2
[T](X) = 2.X + 2
[plus](X1,X2) = X1 + X2
[0] = 0

Problem 1: 

Reduction Order Processor:
-> Rules:
 T(T(x)) -> T(x)
 T(plus(T(y),x)) -> plus(T(plus(x,y)),T(y))
 plus(T(plus(x,y)),x) -> T(plus(x,y))
 plus(T(x),x) -> T(x)
 plus(x,0) -> x
 plus(x,x) -> x
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[L](X) = 2.X
[T](X) = 2.X + 2
[plus](X1,X2) = X1 + X2
[0] = 0

Problem 1: 

Reduction Order Processor:
-> Rules:
 T(plus(T(y),x)) -> plus(T(plus(x,y)),T(y))
 plus(T(plus(x,y)),x) -> T(plus(x,y))
 plus(T(x),x) -> T(x)
 plus(x,0) -> x
 plus(x,x) -> x
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[L](X) = 2.X
[T](X) = 2.X + 2
[plus](X1,X2) = X1 + X2
[0] = 0

Problem 1: 

Reduction Order Processor:
-> Rules:
 plus(T(plus(x,y)),x) -> T(plus(x,y))
 plus(T(x),x) -> T(x)
 plus(x,0) -> x
 plus(x,x) -> x
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[L](X) = 2.X
[T](X) = 2.X + 2
[plus](X1,X2) = X1 + X2 + 2
[0] = 0

Problem 1: 

Reduction Order Processor:
-> Rules:
 plus(T(x),x) -> T(x)
 plus(x,0) -> x
 plus(x,x) -> x
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[L](X) = 2.X
[T](X) = 2.X + 2
[plus](X1,X2) = X1 + X2 + 2
[0] = 0

Problem 1: 

Reduction Order Processor:
-> Rules:
 plus(x,0) -> x
 plus(x,x) -> x
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[L](X) = 2.X
[T](X) = 2.X
[plus](X1,X2) = X1 + X2 + 2
[0] = 2

Problem 1: 

Reduction Order Processor:
-> Rules:
 plus(x,x) -> x
->Interpretation type:
 Linear
->Coefficients:
 Natural Numbers
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[L](X) = 2.X
[T](X) = 2.X
[plus](X1,X2) = X1 + X2 + 2
[0] = 0

Problem 1: 

Dependency Pairs Processor:
-> FAxioms:
 PLUS(plus(x2,x3),x4) = PLUS(x2,plus(x3,x4))
 PLUS(x2,x3) = PLUS(x3,x2)
-> Pairs:
 Empty
-> EAxioms:
 plus(plus(x2,x3),x4) = plus(x2,plus(x3,x4))
 plus(x2,x3) = plus(x3,x2)
-> Rules:
 Empty
-> SRules:
 PLUS(plus(x2,x3),x4) -> PLUS(x2,x3)
 PLUS(x2,plus(x3,x4)) -> PLUS(x3,x4)

Problem 1: 

SCC Processor:
-> FAxioms:
 PLUS(plus(x2,x3),x4) = PLUS(x2,plus(x3,x4))
 PLUS(x2,x3) = PLUS(x3,x2)
-> Pairs:
 Empty
-> EAxioms:
 plus(plus(x2,x3),x4) = plus(x2,plus(x3,x4))
 plus(x2,x3) = plus(x3,x2)
-> Rules:
 Empty
-> SRules:
 PLUS(plus(x2,x3),x4) -> PLUS(x2,x3)
 PLUS(x2,plus(x3,x4)) -> PLUS(x3,x4)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.