YES

Problem:
 a() -> b()
 a() -> c()
 a() -> e()
 b() -> d()
 c() -> a()
 d() -> a()
 d() -> e()
 g(x) -> h(a())
 h(x) -> e()

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  a() -> e()
  d() -> e()
  g(x) -> h(a())
  h(x) -> e()
  b() -> e()
  c() -> e()
  a() -> b()
  a() -> c()
  b() -> d()
  c() -> a()
  d() -> a()
  
   T' = (P union S) with
  
   TRS P:a() -> b()
         a() -> c()
         b() -> d()
         c() -> a()
         d() -> a()
  
   TRS S:a() -> e()
         d() -> e()
         g(x) -> h(a())
         h(x) -> e()
         b() -> e()
         c() -> e()
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  
   CP(S,P union P^-1) = 
  e() = b(), e() = c(), e() = d(), e() = a()
  
   CP(P union P^-1,S) = 
  b() = e(), c() = e(), d() = e(), a() = e()
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [d] = 0,
    
    [b] = 0,
    
    [e] = 0,
    
    [g](x0) = 4x0 + 6,
    
    [a] = 4,
    
    [h](x0) = x0 + 2,
    
    [c] = 4
   orientation:
    a() = 4 >= 0 = e()
    
    d() = 0 >= 0 = e()
    
    g(x) = 4x + 6 >= 6 = h(a())
    
    h(x) = x + 2 >= 0 = e()
    
    b() = 0 >= 0 = e()
    
    c() = 4 >= 0 = e()
   problem:
    d() -> e()
    g(x) -> h(a())
    b() -> e()
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [d] = 4,
     
     [b] = 0,
     
     [e] = 0,
     
     [g](x0) = 2x0 + 6,
     
     [a] = 4,
     
     [h](x0) = x0 + 1
    orientation:
     d() = 4 >= 0 = e()
     
     g(x) = 2x + 6 >= 5 = h(a())
     
     b() = 0 >= 0 = e()
    problem:
     b() -> e()
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [b] = 1,
      
      [e] = 0
     orientation:
      b() = 1 >= 0 = e()
     problem:
      
     Qed