YES

Problem:
 f(x,x) -> f(a(),b())
 b() -> c()

Proof:
 DP Processor:
  DPs:
   f#(x,x) -> b#()
   f#(x,x) -> f#(a(),b())
  TRS:
   f(x,x) -> f(a(),b())
   b() -> c()
  EDG Processor:
   DPs:
    f#(x,x) -> b#()
    f#(x,x) -> f#(a(),b())
   TRS:
    f(x,x) -> f(a(),b())
    b() -> c()
   graph:
    f#(x,x) -> f#(a(),b()) -> f#(x,x) -> b#()
    f#(x,x) -> f#(a(),b()) -> f#(x,x) -> f#(a(),b())
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 2/4
    DPs:
     f#(x,x) -> f#(a(),b())
    TRS:
     f(x,x) -> f(a(),b())
     b() -> c()
    Usable Rule Processor:
     DPs:
      f#(x,x) -> f#(a(),b())
     TRS:
      b() -> c()
     Bounds Processor:
      bound: 2
      enrichment: match
      automaton:
       final states: {4}
       transitions:
        f{#,1}(1,2) -> 4*
        f{#,1}(1,9) -> 4*
        a1() -> 1*
        b1() -> 2*
        c2() -> 3,9*,2
        f{#,0}(3,1) -> 4*
        f{#,0}(3,3) -> 4*
        f{#,0}(3,9) -> 4*
        f{#,0}(9,2) -> 4*
        f{#,0}(2,1) -> 4*
        f{#,0}(2,3) -> 4*
        f{#,0}(2,9) -> 4*
        f{#,0}(3,2) -> 4*
        f{#,0}(9,1) -> 4*
        f{#,0}(9,3) -> 4*
        f{#,0}(9,9) -> 4*
        f{#,0}(1,1) -> 4*
        f{#,0}(1,3) -> 4*
        f{#,0}(1,9) -> 4*
        f{#,0}(2,2) -> 4*
      problem:
       DPs:
        
       TRS:
        b() -> c()
      Qed