/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(0(x1)) -> s(0(x1))
 d(0(x1)) -> 0(x1)
 d(s(x1)) -> s(s(d(p(s(x1)))))
 f(s(x1)) -> d(f(p(s(x1))))
 p(s(x1)) -> x1

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [p](x0) = x0,
   
   [d](x0) = x0,
   
   [s](x0) = x0,
   
   [f](x0) = x0 + 8,
   
   [0](x0) = 2x0 + 14
  orientation:
   f(0(x1)) = 2x1 + 22 >= 2x1 + 14 = s(0(x1))
   
   d(0(x1)) = 2x1 + 14 >= 2x1 + 14 = 0(x1)
   
   d(s(x1)) = x1 >= x1 = s(s(d(p(s(x1)))))
   
   f(s(x1)) = x1 + 8 >= x1 + 8 = d(f(p(s(x1))))
   
   p(s(x1)) = x1 >= x1 = x1
  problem:
   d(0(x1)) -> 0(x1)
   d(s(x1)) -> s(s(d(p(s(x1)))))
   f(s(x1)) -> d(f(p(s(x1))))
   p(s(x1)) -> x1
  Bounds Processor:
   bound: 0
   enrichment: match
   automaton:
    final states: {2,8,3,1}
    transitions:
     f50() -> 2*
     00(2) -> 1*
     s0(7) -> 3*
     s0(2) -> 4*
     s0(6) -> 7*
     d0(5) -> 6*
     d0(9) -> 8*
     p0(4) -> 5*
     f0(5) -> 9*
     1 -> 6*
     2 -> 5*
     3 -> 6*
     8 -> 9*
   problem:
    
   Qed