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


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YES

Problem:
 P(x1) -> Q(Q(p(x1)))
 p(p(x1)) -> q(q(x1))
 p(Q(Q(x1))) -> Q(Q(p(x1)))
 Q(p(q(x1))) -> q(p(Q(x1)))
 q(q(p(x1))) -> p(q(q(x1)))
 q(Q(x1)) -> x1
 Q(q(x1)) -> x1
 p(P(x1)) -> x1
 P(p(x1)) -> x1

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [p](x0) = 2x0 + 1,
   
   [q](x0) = 2x0 + 1,
   
   [P](x0) = 2x0 + 1,
   
   [Q](x0) = x0
  orientation:
   P(x1) = 2x1 + 1 >= 2x1 + 1 = Q(Q(p(x1)))
   
   p(p(x1)) = 4x1 + 3 >= 4x1 + 3 = q(q(x1))
   
   p(Q(Q(x1))) = 2x1 + 1 >= 2x1 + 1 = Q(Q(p(x1)))
   
   Q(p(q(x1))) = 4x1 + 3 >= 4x1 + 3 = q(p(Q(x1)))
   
   q(q(p(x1))) = 8x1 + 7 >= 8x1 + 7 = p(q(q(x1)))
   
   q(Q(x1)) = 2x1 + 1 >= x1 = x1
   
   Q(q(x1)) = 2x1 + 1 >= x1 = x1
   
   p(P(x1)) = 4x1 + 3 >= x1 = x1
   
   P(p(x1)) = 4x1 + 3 >= x1 = x1
  problem:
   P(x1) -> Q(Q(p(x1)))
   p(p(x1)) -> q(q(x1))
   p(Q(Q(x1))) -> Q(Q(p(x1)))
   Q(p(q(x1))) -> q(p(Q(x1)))
   q(q(p(x1))) -> p(q(q(x1)))
  String Reversal Processor:
   P(x1) -> p(Q(Q(x1)))
   p(p(x1)) -> q(q(x1))
   Q(Q(p(x1))) -> p(Q(Q(x1)))
   q(p(Q(x1))) -> Q(p(q(x1)))
   p(q(q(x1))) -> q(q(p(x1)))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [p](x0) = 2x0,
     
     [q](x0) = 2x0,
     
     [P](x0) = 11x0 + 1,
     
     [Q](x0) = 2x0
    orientation:
     P(x1) = 11x1 + 1 >= 8x1 = p(Q(Q(x1)))
     
     p(p(x1)) = 4x1 >= 4x1 = q(q(x1))
     
     Q(Q(p(x1))) = 8x1 >= 8x1 = p(Q(Q(x1)))
     
     q(p(Q(x1))) = 8x1 >= 8x1 = Q(p(q(x1)))
     
     p(q(q(x1))) = 8x1 >= 8x1 = q(q(p(x1)))
    problem:
     p(p(x1)) -> q(q(x1))
     Q(Q(p(x1))) -> p(Q(Q(x1)))
     q(p(Q(x1))) -> Q(p(q(x1)))
     p(q(q(x1))) -> q(q(p(x1)))
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [p](x0) = 2x0 + 8,
      
      [q](x0) = x0,
      
      [Q](x0) = x0
     orientation:
      p(p(x1)) = 4x1 + 24 >= x1 = q(q(x1))
      
      Q(Q(p(x1))) = 2x1 + 8 >= 2x1 + 8 = p(Q(Q(x1)))
      
      q(p(Q(x1))) = 2x1 + 8 >= 2x1 + 8 = Q(p(q(x1)))
      
      p(q(q(x1))) = 2x1 + 8 >= 2x1 + 8 = q(q(p(x1)))
     problem:
      Q(Q(p(x1))) -> p(Q(Q(x1)))
      q(p(Q(x1))) -> Q(p(q(x1)))
      p(q(q(x1))) -> q(q(p(x1)))
     String Reversal Processor:
      p(Q(Q(x1))) -> Q(Q(p(x1)))
      Q(p(q(x1))) -> q(p(Q(x1)))
      q(q(p(x1))) -> p(q(q(x1)))
      Bounds Processor:
       bound: 0
       enrichment: match
       automaton:
        final states: {8,5,1}
        transitions:
         Q0(4) -> 1*
         Q0(3) -> 4*
         Q0(2) -> 6*
         q0(7) -> 5*
         q0(9) -> 10*
         q0(2) -> 9*
         f40() -> 2*
         p0(6) -> 7*
         p0(2) -> 3*
         p0(10) -> 8*
         8 -> 9,10
         1 -> 3,7
         5 -> 6,4
       problem:
        
       Qed