YES

Problem 1: 

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

Problem 1: 

Innermost Equivalent Processor:
-> Rules:
 f(x:S,f(a,a)) -> f(f(a,f(f(a,a),a)),x:S)
-> 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(a,f(f(a,a),a)),x:S)
-> Rules:
 f(x:S,f(a,a)) -> f(f(a,f(f(a,a),a)),x:S)

Problem 1: 

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

Problem 1: 

Reduction Pairs Processor:
-> Pairs:
 F(x:S,f(a,a)) -> F(f(a,f(f(a,a),a)),x:S)
-> Rules:
 f(x:S,f(a,a)) -> f(f(a,f(f(a,a),a)),x:S)
-> Usable rules:
 f(x:S,f(a,a)) -> f(f(a,f(f(a,a),a)),x:S)
->Interpretation type:
 Linear
->Coefficients:
 All rationals
->Dimension:
 1
->Bound:
 4
->Interpretation:
 
[f](X1,X2) = 1/3.X1 + 1/3.X2
[a] = 2
[fSNonEmpty] = 0
[F](X1,X2) = 1/4.X1 + 1/4.X2

Problem 1: 

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

The problem is finite.