We consider the following Problem:

  Strict Trs: {f(0(), 1(), x) -> f(x, x, x)}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(O(1),O(1))

Proof:
  We consider the following Problem:
  
    Strict Trs: {f(0(), 1(), x) -> f(x, x, x)}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(O(1),O(1))
  
  Proof:
    We consider the following Problem:
    
      Strict Trs: {f(0(), 1(), x) -> f(x, x, x)}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(O(1),O(1))
    
    Proof:
      We have computed the following dependency pairs
      
        Strict DPs: {f^#(0(), 1(), x) -> f^#(x, x, x)}
      
      We consider the following Problem:
      
        Strict DPs: {f^#(0(), 1(), x) -> f^#(x, x, x)}
        Strict Trs: {f(0(), 1(), x) -> f(x, x, x)}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(O(1),O(1))
      
      Proof:
        No rule is usable.
        
        We consider the following Problem:
        
          Strict DPs: {f^#(0(), 1(), x) -> f^#(x, x, x)}
          StartTerms: basic terms
          Strategy: innermost
        
        Certificate: YES(O(1),O(1))
        
        Proof:
          We consider the following Problem:
          
            Strict DPs: {f^#(0(), 1(), x) -> f^#(x, x, x)}
            StartTerms: basic terms
            Strategy: innermost
          
          Certificate: YES(O(1),O(1))
          
          Proof:
            We consider the following Problem:
            
              Strict DPs: {f^#(0(), 1(), x) -> f^#(x, x, x)}
              StartTerms: basic terms
              Strategy: innermost
            
            Certificate: YES(O(1),O(1))
            
            Proof:
              We consider the the dependency-graph
              
                1: f^#(0(), 1(), x) -> f^#(x, x, x)
                
              
              together with the congruence-graph
              
                ->1:{1}                                                     Noncyclic, trivial, SCC
                
                
                Here dependency-pairs are as follows:
                
                Strict DPs:
                  {1: f^#(0(), 1(), x) -> f^#(x, x, x)}
              
              The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
              
                {1: f^#(0(), 1(), x) -> f^#(x, x, x)}
              
              We consider the following Problem:
              
                StartTerms: basic terms
                Strategy: innermost
              
              Certificate: YES(O(1),O(1))
              
              Proof:
                We consider the following Problem:
                
                  StartTerms: basic terms
                  Strategy: innermost
                
                Certificate: YES(O(1),O(1))
                
                Proof:
                  Empty rules are trivially bounded

Hurray, we answered YES(O(1),O(1))