YES

Problem 1: 

(THEORY 
(AC plus))
(RULES
f(plus(a,a)) -> plus(f(a),f(a))
plus(a,f(f(a))) -> plus(f(a),f(a))
)

Problem 1: 

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

Problem 1: 

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

Problem 1: 

Dependency Pairs Processor:
-> FAxioms:
 PLUS(plus(x0,x1),x2) = PLUS(x0,plus(x1,x2))
 PLUS(x0,x1) = PLUS(x1,x0)
-> Pairs:
 Empty
-> EAxioms:
 plus(plus(x0,x1),x2) = plus(x0,plus(x1,x2))
 plus(x0,x1) = plus(x1,x0)
-> Rules:
 Empty
-> SRules:
 PLUS(plus(x0,x1),x2) -> PLUS(x0,x1)
 PLUS(x0,plus(x1,x2)) -> PLUS(x1,x2)

Problem 1: 

SCC Processor:
-> FAxioms:
 PLUS(plus(x0,x1),x2) = PLUS(x0,plus(x1,x2))
 PLUS(x0,x1) = PLUS(x1,x0)
-> Pairs:
 Empty
-> EAxioms:
 plus(plus(x0,x1),x2) = plus(x0,plus(x1,x2))
 plus(x0,x1) = plus(x1,x0)
-> Rules:
 Empty
-> SRules:
 PLUS(plus(x0,x1),x2) -> PLUS(x0,x1)
 PLUS(x0,plus(x1,x2)) -> PLUS(x1,x2)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.