YES

Problem 1: 

(VAR x)
(THEORY 
(AC times))
(RULES
fac(0) -> s(0)
fac(s(x)) -> times(s(x),fac(p(s(x))))
p(s(x)) -> x
)

Problem 1: 

Dependency Pairs Processor:
-> FAxioms:
 TIMES(times(x1,x2),x3) = TIMES(x1,times(x2,x3))
 TIMES(x1,x2) = TIMES(x2,x1)
-> Pairs:
 FAC(s(x)) -> FAC(p(s(x)))
 FAC(s(x)) -> P(s(x))
-> EAxioms:
 times(times(x1,x2),x3) = times(x1,times(x2,x3))
 times(x1,x2) = times(x2,x1)
-> Rules:
 fac(0) -> s(0)
 fac(s(x)) -> times(s(x),fac(p(s(x))))
 p(s(x)) -> x
-> SRules:
 TIMES(times(x1,x2),x3) -> TIMES(x1,x2)
 TIMES(x1,times(x2,x3)) -> TIMES(x2,x3)

Problem 1: 

SCC Processor:
-> FAxioms:
 TIMES(times(x1,x2),x3) = TIMES(x1,times(x2,x3))
 TIMES(x1,x2) = TIMES(x2,x1)
-> Pairs:
 FAC(s(x)) -> FAC(p(s(x)))
 FAC(s(x)) -> P(s(x))
-> EAxioms:
 times(times(x1,x2),x3) = times(x1,times(x2,x3))
 times(x1,x2) = times(x2,x1)
-> Rules:
 fac(0) -> s(0)
 fac(s(x)) -> times(s(x),fac(p(s(x))))
 p(s(x)) -> x
-> SRules:
 TIMES(times(x1,x2),x3) -> TIMES(x1,x2)
 TIMES(x1,times(x2,x3)) -> TIMES(x2,x3)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 FAC(s(x)) -> FAC(p(s(x)))
-> FAxioms:
 times(times(x1,x2),x3) -> times(x1,times(x2,x3))
 times(x1,x2) -> times(x2,x1)
-> EAxioms:
 times(times(x1,x2),x3) = times(x1,times(x2,x3))
 times(x1,x2) = times(x2,x1)
->->-> Rules:
 fac(0) -> s(0)
 fac(s(x)) -> times(s(x),fac(p(s(x))))
 p(s(x)) -> x
-> SRules:
 Empty

Problem 1: 

Reduction Pairs Processor:
-> FAxioms:
 Empty
-> Pairs:
 FAC(s(x)) -> FAC(p(s(x)))
-> EAxioms:
 times(times(x1,x2),x3) = times(x1,times(x2,x3))
 times(x1,x2) = times(x2,x1)
-> Usable Equations:
 Empty
-> Rules:
 fac(0) -> s(0)
 fac(s(x)) -> times(s(x),fac(p(s(x))))
 p(s(x)) -> x
-> Usable Rules:
 p(s(x)) -> x
-> SRules:
 Empty
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 3
->Interpretation:
 
[fac](X) = 0
[p](X) = 1/2.X
[0] = 0
[s](X) = 2.X + 3
[times](X1,X2) = 0
[FAC](X) = 3/2.X
[P](X) = 0
[TIMES](X1,X2) = 0

Problem 1: 

SCC Processor:
-> FAxioms:
 Empty
-> Pairs:
 Empty
-> EAxioms:
 times(times(x1,x2),x3) = times(x1,times(x2,x3))
 times(x1,x2) = times(x2,x1)
-> Rules:
 fac(0) -> s(0)
 fac(s(x)) -> times(s(x),fac(p(s(x))))
 p(s(x)) -> x
-> SRules:
 Empty
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.