YES

Problem 1: 

(VAR v_NonEmpty:S x:S)
(RULES
f(x:S,f(a,a)) -> f(f(f(f(a,a),a),x:S),a)
)

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 f(x:S,f(a,a)) -> f(f(f(f(a,a),a),x:S),a)
-> The term rewriting system is non-overlaping or locally confluent overlay system. Therefore, innermost termination implies termination.
 

Problem 1: 

Dependency Pairs Processor:
-> Pairs:
 F(x:S,f(a,a)) -> F(f(f(f(a,a),a),x:S),a)
 F(x:S,f(a,a)) -> F(f(f(a,a),a),x:S)
-> Rules:
 f(x:S,f(a,a)) -> f(f(f(f(a,a),a),x:S),a)

Problem 1: 

SCC Processor:
-> Pairs:
 F(x:S,f(a,a)) -> F(f(f(f(a,a),a),x:S),a)
 F(x:S,f(a,a)) -> F(f(f(a,a),a),x:S)
-> Rules:
 f(x:S,f(a,a)) -> f(f(f(f(a,a),a),x:S),a)
->Strongly Connected Components:
->->Cycle:
->->-> Pairs:
 F(x:S,f(a,a)) -> F(f(f(a,a),a),x:S)
->->-> Rules:
 f(x:S,f(a,a)) -> f(f(f(f(a,a),a),x:S),a)

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 F(x:S,f(a,a)) -> F(f(f(a,a),a),x:S)
-> Rules:
 f(x:S,f(a,a)) -> f(f(f(f(a,a),a),x:S),a)
-> Usable rules:
 f(x:S,f(a,a)) -> f(f(f(f(a,a),a),x:S),a)
->Interpretation type:
 Simple mixed
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 2
->Interpretation:
 
[f](X1,X2) = X1.X2
[a] = 1/2
[fSNonEmpty] = 0
[F](X1,X2) = X1.X2 + 2.X1 + 2.X2

Problem 1: 

SCC Processor:
-> Pairs:
 Empty
-> Rules:
 f(x:S,f(a,a)) -> f(f(f(f(a,a),a),x:S),a)
->Strongly Connected Components:
 There is no strongly connected component

The problem is finite.