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()
  TDG 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) -> f#(a(),b())
    f#(x,x) -> f#(a(),b()) -> f#(x,x) -> 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()
    Bounds Processor:
     bound: 0
     enrichment: match-dp
     automaton:
      final states: {1}
      transitions:
       f{#,0}(3,2) -> 1*
       a0() -> 3*
       c0() -> 2*
       b0() -> 2*
     problem:
      DPs:
       
      TRS:
       f(x,x) -> f(a(),b())
       b() -> c()
     Qed