/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(a(x1)) -> b(b(b(x1)))
 b(x1) -> c(c(d(x1)))
 c(x1) -> d(d(d(x1)))
 b(c(x1)) -> c(b(x1))
 b(c(d(x1))) -> a(x1)

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = x0 + 2,
   
   [d](x0) = x0,
   
   [b](x0) = x0 + 4,
   
   [a](x0) = x0 + 6
  orientation:
   a(a(x1)) = x1 + 12 >= x1 + 12 = b(b(b(x1)))
   
   b(x1) = x1 + 4 >= x1 + 4 = c(c(d(x1)))
   
   c(x1) = x1 + 2 >= x1 = d(d(d(x1)))
   
   b(c(x1)) = x1 + 6 >= x1 + 6 = c(b(x1))
   
   b(c(d(x1))) = x1 + 6 >= x1 + 6 = a(x1)
  problem:
   a(a(x1)) -> b(b(b(x1)))
   b(x1) -> c(c(d(x1)))
   b(c(x1)) -> c(b(x1))
   b(c(d(x1))) -> a(x1)
  String Reversal Processor:
   a(a(x1)) -> b(b(b(x1)))
   b(x1) -> d(c(c(x1)))
   c(b(x1)) -> b(c(x1))
   d(c(b(x1))) -> a(x1)
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [c](x0) = x0,
     
     [d](x0) = x0 + 1,
     
     [b](x0) = x0 + 1,
     
     [a](x0) = x0 + 2
    orientation:
     a(a(x1)) = x1 + 4 >= x1 + 3 = b(b(b(x1)))
     
     b(x1) = x1 + 1 >= x1 + 1 = d(c(c(x1)))
     
     c(b(x1)) = x1 + 1 >= x1 + 1 = b(c(x1))
     
     d(c(b(x1))) = x1 + 2 >= x1 + 2 = a(x1)
    problem:
     b(x1) -> d(c(c(x1)))
     c(b(x1)) -> b(c(x1))
     d(c(b(x1))) -> a(x1)
    KBO Processor:
     weight function:
      w0 = 1
      w(d) = w(b) = w(a) = 1
      w(c) = 0
     precedence:
      c > b > d ~ a
     problem:
      
     Qed