YES

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

Problem 1: 

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

The problem is finite.