/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

Problem:
 a(x1) -> b(c(x1))
 a(b(x1)) -> b(a(x1))
 d(c(x1)) -> d(a(x1))
 a(c(x1)) -> c(a(x1))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 1 0]  
   [d](x0) = [0 1 0]x0
             [0 0 0]  ,
   
             [1 0 0]  
   [b](x0) = [0 0 0]x0
             [0 0 1]  ,
   
             [1 0 1]     [0]
   [c](x0) = [0 1 1]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 1]  
   [a](x0) = [0 1 1]x0
             [0 0 0]  
  orientation:
           [1 0 1]      [1 0 1]             
   a(x1) = [0 1 1]x1 >= [0 0 0]x1 = b(c(x1))
           [0 0 0]      [0 0 0]             
   
              [1 0 1]      [1 0 1]             
   a(b(x1)) = [0 0 1]x1 >= [0 0 0]x1 = b(a(x1))
              [0 0 0]      [0 0 0]             
   
              [1 1 2]     [1]    [1 1 2]             
   d(c(x1)) = [0 1 1]x1 + [1] >= [0 1 1]x1 = d(a(x1))
              [0 0 0]     [0]    [0 0 0]             
   
              [1 0 1]     [0]    [1 0 1]     [0]           
   a(c(x1)) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = c(a(x1))
              [0 0 0]     [0]    [0 0 0]     [0]           
  problem:
   a(x1) -> b(c(x1))
   a(b(x1)) -> b(a(x1))
   a(c(x1)) -> c(a(x1))
  String Reversal Processor:
   a(x1) -> c(b(x1))
   b(a(x1)) -> a(b(x1))
   c(a(x1)) -> a(c(x1))
   Matrix Interpretation Processor: dim=4
    
    interpretation:
               [1 0 0 0]     [0]
               [1 0 0 1]     [1]
     [b](x0) = [1 0 0 0]x0 + [1]
               [0 0 0 0]     [0],
     
               [1 0 0 0]  
               [0 0 0 0]  
     [c](x0) = [0 0 0 0]x0
               [0 0 0 0]  ,
     
               [1 0 0 0]     [1]
               [0 0 0 0]     [0]
     [a](x0) = [0 0 1 0]x0 + [0]
               [0 0 0 1]     [0]
    orientation:
             [1 0 0 0]     [1]    [1 0 0 0]             
             [0 0 0 0]     [0]    [0 0 0 0]             
     a(x1) = [0 0 1 0]x1 + [0] >= [0 0 0 0]x1 = c(b(x1))
             [0 0 0 1]     [0]    [0 0 0 0]             
     
                [1 0 0 0]     [1]    [1 0 0 0]     [1]           
                [1 0 0 1]     [2]    [0 0 0 0]     [0]           
     b(a(x1)) = [1 0 0 0]x1 + [2] >= [1 0 0 0]x1 + [1] = a(b(x1))
                [0 0 0 0]     [0]    [0 0 0 0]     [0]           
     
                [1 0 0 0]     [1]    [1 0 0 0]     [1]           
                [0 0 0 0]     [0]    [0 0 0 0]     [0]           
     c(a(x1)) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = a(c(x1))
                [0 0 0 0]     [0]    [0 0 0 0]     [0]           
    problem:
     b(a(x1)) -> a(b(x1))
     c(a(x1)) -> a(c(x1))
    String Reversal Processor:
     a(b(x1)) -> b(a(x1))
     a(c(x1)) -> c(a(x1))
     KBO Processor:
      weight function:
       w0 = 1
       w(b) = w(c) = 1
       w(a) = 0
      precedence:
       a > b ~ c
      problem:
       
      Qed