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


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YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> x1
  b(a(x1)) -> c(b(x1))
  c(a(x1)) -> a(a(b(c(x1))))
  DP Processor:
   DPs:
    b#(a(x1)) -> b#(x1)
    b#(a(x1)) -> c#(b(x1))
    c#(a(x1)) -> c#(x1)
    c#(a(x1)) -> b#(c(x1))
   TRS:
    b(a(x1)) -> x1
    b(a(x1)) -> c(b(x1))
    c(a(x1)) -> a(a(b(c(x1))))
   TDG Processor:
    DPs:
     b#(a(x1)) -> b#(x1)
     b#(a(x1)) -> c#(b(x1))
     c#(a(x1)) -> c#(x1)
     c#(a(x1)) -> b#(c(x1))
    TRS:
     b(a(x1)) -> x1
     b(a(x1)) -> c(b(x1))
     c(a(x1)) -> a(a(b(c(x1))))
    graph:
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> b#(c(x1))
     c#(a(x1)) -> c#(x1) -> c#(a(x1)) -> c#(x1)
     c#(a(x1)) -> b#(c(x1)) -> b#(a(x1)) -> c#(b(x1))
     c#(a(x1)) -> b#(c(x1)) -> b#(a(x1)) -> b#(x1)
     b#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> b#(c(x1))
     b#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> c#(x1)
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> c#(b(x1))
     b#(a(x1)) -> b#(x1) -> b#(a(x1)) -> b#(x1)
    Arctic Interpretation Processor:
     dimension: 2
     usable rules:
      b(a(x1)) -> x1
      b(a(x1)) -> c(b(x1))
      c(a(x1)) -> a(a(b(c(x1))))
     interpretation:
      [c#](x0) = [0 1]x0 + [0],
      
      [b#](x0) = [-& 0 ]x0 + [0],
      
                [-& -&]     [0 ]
      [c](x0) = [-& 1 ]x0 + [-&],
      
                [-& -&]     [0]
      [a](x0) = [1  1 ]x0 + [3],
      
                [0  -1]     [-4]
      [b](x0) = [-& -1]x0 + [-1]
     orientation:
      b#(a(x1)) = [1 1]x1 + [3] >= [-& 0 ]x1 + [0] = b#(x1)
      
      b#(a(x1)) = [1 1]x1 + [3] >= [0 0]x1 + [0] = c#(b(x1))
      
      c#(a(x1)) = [2 2]x1 + [4] >= [0 1]x1 + [0] = c#(x1)
      
      c#(a(x1)) = [2 2]x1 + [4] >= [-& 1 ]x1 + [0] = b#(c(x1))
      
                 [0 0]     [2]           
      b(a(x1)) = [0 0]x1 + [2] >= x1 = x1
      
                 [0 0]     [2]    [-& -&]     [0]           
      b(a(x1)) = [0 0]x1 + [2] >= [-& 0 ]x1 + [0] = c(b(x1))
      
                 [-& -&]     [0]    [-& -&]     [0]                 
      c(a(x1)) = [2  2 ]x1 + [4] >= [-& 2 ]x1 + [4] = a(a(b(c(x1))))
     problem:
      DPs:
       
      TRS:
       b(a(x1)) -> x1
       b(a(x1)) -> c(b(x1))
       c(a(x1)) -> a(a(b(c(x1))))
     Qed