/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:
 t(f(x1)) -> t(c(n(x1)))
 n(f(x1)) -> f(n(x1))
 o(f(x1)) -> f(o(x1))
 n(s(x1)) -> f(s(x1))
 o(s(x1)) -> f(s(x1))
 c(f(x1)) -> f(c(x1))
 c(n(x1)) -> n(c(x1))
 c(o(x1)) -> o(c(x1))
 c(o(x1)) -> o(x1)

Proof:
 String Reversal Processor:
  f(t(x1)) -> n(c(t(x1)))
  f(n(x1)) -> n(f(x1))
  f(o(x1)) -> o(f(x1))
  s(n(x1)) -> s(f(x1))
  s(o(x1)) -> s(f(x1))
  f(c(x1)) -> c(f(x1))
  n(c(x1)) -> c(n(x1))
  o(c(x1)) -> c(o(x1))
  o(c(x1)) -> o(x1)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [s](x0) = x0 + 8,
    
    [o](x0) = 4x0 + 8,
    
    [c](x0) = x0,
    
    [n](x0) = 2x0 + 2,
    
    [t](x0) = 8x0,
    
    [f](x0) = 2x0 + 2
   orientation:
    f(t(x1)) = 16x1 + 2 >= 16x1 + 2 = n(c(t(x1)))
    
    f(n(x1)) = 4x1 + 6 >= 4x1 + 6 = n(f(x1))
    
    f(o(x1)) = 8x1 + 18 >= 8x1 + 16 = o(f(x1))
    
    s(n(x1)) = 2x1 + 10 >= 2x1 + 10 = s(f(x1))
    
    s(o(x1)) = 4x1 + 16 >= 2x1 + 10 = s(f(x1))
    
    f(c(x1)) = 2x1 + 2 >= 2x1 + 2 = c(f(x1))
    
    n(c(x1)) = 2x1 + 2 >= 2x1 + 2 = c(n(x1))
    
    o(c(x1)) = 4x1 + 8 >= 4x1 + 8 = c(o(x1))
    
    o(c(x1)) = 4x1 + 8 >= 4x1 + 8 = o(x1)
   problem:
    f(t(x1)) -> n(c(t(x1)))
    f(n(x1)) -> n(f(x1))
    s(n(x1)) -> s(f(x1))
    f(c(x1)) -> c(f(x1))
    n(c(x1)) -> c(n(x1))
    o(c(x1)) -> c(o(x1))
    o(c(x1)) -> o(x1)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
               [1 0 1]     [0]
     [s](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [1],
     
               [1 0 0]     [0]
     [o](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [1],
     
               [1 0 0]  
     [c](x0) = [0 0 0]x0
               [0 0 0]  ,
     
               [1 0 0]     [0]
     [n](x0) = [0 1 1]x0 + [0]
               [0 1 1]     [1],
     
               [1 0 0]     [0]
     [t](x0) = [0 0 0]x0 + [1]
               [1 0 0]     [1],
     
               [1 0 0]     [1]
     [f](x0) = [0 1 1]x0 + [0]
               [0 1 1]     [0]
    orientation:
                [1 0 0]     [1]    [1 0 0]     [0]              
     f(t(x1)) = [1 0 0]x1 + [2] >= [0 0 0]x1 + [0] = n(c(t(x1)))
                [1 0 0]     [2]    [0 0 0]     [1]              
     
                [1 0 0]     [1]    [1 0 0]     [1]           
     f(n(x1)) = [0 2 2]x1 + [1] >= [0 2 2]x1 + [0] = n(f(x1))
                [0 2 2]     [1]    [0 2 2]     [1]           
     
                [1 1 1]     [1]    [1 1 1]     [1]           
     s(n(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = s(f(x1))
                [0 0 0]     [1]    [0 0 0]     [1]           
     
                [1 0 0]     [1]    [1 0 0]     [1]           
     f(c(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = c(f(x1))
                [0 0 0]     [0]    [0 0 0]     [0]           
     
                [1 0 0]     [0]    [1 0 0]             
     n(c(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = c(n(x1))
                [0 0 0]     [1]    [0 0 0]             
     
                [1 0 0]     [0]    [1 0 0]             
     o(c(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 = c(o(x1))
                [0 0 0]     [1]    [0 0 0]             
     
                [1 0 0]     [0]    [1 0 0]     [0]        
     o(c(x1)) = [0 0 0]x1 + [0] >= [0 0 0]x1 + [0] = o(x1)
                [0 0 0]     [1]    [0 0 0]     [1]        
    problem:
     f(n(x1)) -> n(f(x1))
     s(n(x1)) -> s(f(x1))
     f(c(x1)) -> c(f(x1))
     n(c(x1)) -> c(n(x1))
     o(c(x1)) -> c(o(x1))
     o(c(x1)) -> o(x1)
    KBO Processor:
     weight function:
      w0 = 1
      w(s) = w(o) = w(c) = w(n) = 1
      w(f) = 0
     precedence:
      f > n > o > s ~ c
     problem:
      
     Qed