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

Proof:
 String Reversal Processor:
  f(f(x1)) -> b(b(b(x1)))
  f(a(x1)) -> a(a(f(x1)))
  b(b(x1)) -> c(a(c(c(x1))))
  b(d(x1)) -> b(a(d(x1)))
  c(c(x1)) -> d(d(d(x1)))
  d(b(x1)) -> b(d(x1))
  d(d(c(x1))) -> f(x1)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [d](x0) = x0 + 4,
    
    [c](x0) = x0 + 6,
    
    [a](x0) = x0,
    
    [b](x0) = x0 + 9,
    
    [f](x0) = x0 + 14
   orientation:
    f(f(x1)) = x1 + 28 >= x1 + 27 = b(b(b(x1)))
    
    f(a(x1)) = x1 + 14 >= x1 + 14 = a(a(f(x1)))
    
    b(b(x1)) = x1 + 18 >= x1 + 18 = c(a(c(c(x1))))
    
    b(d(x1)) = x1 + 13 >= x1 + 13 = b(a(d(x1)))
    
    c(c(x1)) = x1 + 12 >= x1 + 12 = d(d(d(x1)))
    
    d(b(x1)) = x1 + 13 >= x1 + 13 = b(d(x1))
    
    d(d(c(x1))) = x1 + 14 >= x1 + 14 = f(x1)
   problem:
    f(a(x1)) -> a(a(f(x1)))
    b(b(x1)) -> c(a(c(c(x1))))
    b(d(x1)) -> b(a(d(x1)))
    c(c(x1)) -> d(d(d(x1)))
    d(b(x1)) -> b(d(x1))
    d(d(c(x1))) -> f(x1)
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [d](x0) = x0 + 4,
     
     [c](x0) = x0 + 8,
     
     [a](x0) = x0,
     
     [b](x0) = x0 + 14,
     
     [f](x0) = x0 + 4
    orientation:
     f(a(x1)) = x1 + 4 >= x1 + 4 = a(a(f(x1)))
     
     b(b(x1)) = x1 + 28 >= x1 + 24 = c(a(c(c(x1))))
     
     b(d(x1)) = x1 + 18 >= x1 + 18 = b(a(d(x1)))
     
     c(c(x1)) = x1 + 16 >= x1 + 12 = d(d(d(x1)))
     
     d(b(x1)) = x1 + 18 >= x1 + 18 = b(d(x1))
     
     d(d(c(x1))) = x1 + 16 >= x1 + 4 = f(x1)
    problem:
     f(a(x1)) -> a(a(f(x1)))
     b(d(x1)) -> b(a(d(x1)))
     d(b(x1)) -> b(d(x1))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {8,5,1}
      transitions:
       f50() -> 2*
       a0(4) -> 1*
       a0(6) -> 7*
       a0(3) -> 4*
       f0(2) -> 3*
       b0(7) -> 5*
       b0(6) -> 8*
       d0(2) -> 6*
       b1(15) -> 16*
       a1(14) -> 15*
       d1(13) -> 14*
       1 -> 3*
       2 -> 13*
       8 -> 14,6
       16 -> 8*
     problem:
      
     Qed