YES

Problem:
 g(x,a(),b()) -> g(b(),b(),a())

Proof:
 DP Processor:
  DPs:
   g#(x,a(),b()) -> g#(b(),b(),a())
  TRS:
   g(x,a(),b()) -> g(b(),b(),a())
  Usable Rule Processor:
   DPs:
    g#(x,a(),b()) -> g#(b(),b(),a())
   TRS:
    
   Restore Modifier:
    DPs:
     g#(x,a(),b()) -> g#(b(),b(),a())
    TRS:
     g(x,a(),b()) -> g(b(),b(),a())
    SCC Processor:
     #sccs: 1
     #rules: 1
     #arcs: 1/1
     DPs:
      g#(x,a(),b()) -> g#(b(),b(),a())
     TRS:
      g(x,a(),b()) -> g(b(),b(),a())
     Matrix Interpretation Processor:
      dimension: 1
      interpretation:
       [g#](x0, x1, x2) = x1,
       
       [g](x0, x1, x2) = 0,
       
       [b] = 0,
       
       [a] = 1
      orientation:
       g#(x,a(),b()) = 1 >= 0 = g#(b(),b(),a())
       
       g(x,a(),b()) = 0 >= 0 = g(b(),b(),a())
      problem:
       DPs:
        
       TRS:
        g(x,a(),b()) -> g(b(),b(),a())
      Qed