/export/starexec/sandbox/solver/bin/starexec_run_ttt2 /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> a(a(a(b(x1))))
  a(b(x1)) -> a(a(x1))
  a(a(x1)) -> b(c(a(x1)))
  Matrix Interpretation Processor: dim=2
   
   interpretation:
                   [0]
    [a](x0) = x0 + [1],
    
              [1 2]     [0]
    [b](x0) = [0 3]x0 + [1],
    
              [1 0]  
    [c](x0) = [0 0]x0
   orientation:
               [1 2]     [2]    [1 2]     [0]                 
    b(a(x1)) = [0 3]x1 + [4] >= [0 3]x1 + [4] = a(a(a(b(x1))))
    
               [1 2]     [0]         [0]           
    a(b(x1)) = [0 3]x1 + [2] >= x1 + [2] = a(a(x1))
    
                    [0]    [1 0]     [0]              
    a(a(x1)) = x1 + [2] >= [0 0]x1 + [1] = b(c(a(x1)))
   problem:
    a(b(x1)) -> a(a(x1))
    a(a(x1)) -> b(c(a(x1)))
   String Reversal Processor:
    b(a(x1)) -> a(a(x1))
    a(a(x1)) -> a(c(b(x1)))
    Bounds Processor:
     bound: 1
     enrichment: match
     automaton:
      final states: {4,1}
      transitions:
       c0(5) -> 6*
       a1(9) -> 10*
       b0(2) -> 5*
       a0(3) -> 1*
       a0(6) -> 4*
       a0(2) -> 3*
       f30() -> 2*
       c1(8) -> 9*
       b1(15) -> 16*
       b1(7) -> 8*
       16 -> 8*
       4 -> 3*
       2 -> 7*
       6 -> 15*
       1 -> 8,5
       10 -> 1*
     problem:
      
     Qed