YES

Problem 1: 

(VAR x)
(THEORY 
(AC plus))
(RULES
f(x) -> plus(g(x),a)
plus(f(x),g(x)) -> plus(g(x),plus(g(x),g(x)))
)

Problem 1: 

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

Problem 1: 

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

Problem 1: 

Dependency Pairs Processor:
-> FAxioms:
 PLUS(plus(x1,x2),x3) = PLUS(x1,plus(x2,x3))
 PLUS(x1,x2) = PLUS(x2,x1)
-> Pairs:
 Empty
-> EAxioms:
 plus(plus(x1,x2),x3) = plus(x1,plus(x2,x3))
 plus(x1,x2) = plus(x2,x1)
-> Rules:
 Empty
-> SRules:
 PLUS(plus(x1,x2),x3) -> PLUS(x1,x2)
 PLUS(x1,plus(x2,x3)) -> PLUS(x2,x3)

Problem 1: 

SCC Processor:
-> FAxioms:
 PLUS(plus(x1,x2),x3) = PLUS(x1,plus(x2,x3))
 PLUS(x1,x2) = PLUS(x2,x1)
-> Pairs:
 Empty
-> EAxioms:
 plus(plus(x1,x2),x3) = plus(x1,plus(x2,x3))
 plus(x1,x2) = plus(x2,x1)
-> Rules:
 Empty
-> SRules:
 PLUS(plus(x1,x2),x3) -> PLUS(x1,x2)
 PLUS(x1,plus(x2,x3)) -> PLUS(x2,x3)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.