/export/starexec/sandbox2/solver/bin/starexec_run_ttt2 /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [b](x0) = x0 + 3,
   
   [d](x0) = x0 + 4,
   
   [a](x0) = x0 + 2,
   
   [c](x0) = x0 + 8
  orientation:
   a(a(a(x1))) = x1 + 6 >= x1 + 6 = b(b(x1))
   
   b(b(b(x1))) = x1 + 9 >= x1 + 8 = c(x1)
   
   c(x1) = x1 + 8 >= x1 + 8 = d(d(x1))
   
   d(x1) = x1 + 4 >= x1 + 4 = a(a(x1))
  problem:
   a(a(a(x1))) -> b(b(x1))
   c(x1) -> d(d(x1))
   d(x1) -> a(a(x1))
  WPO Processor:
   algebra: Sum
   weight function:
    w0 = 0
    w(c) = 6
    w(d) = 3
    w(a) = 1
    w(b) = 0
   status function:
    st(d) = st(c) = st(b) = st(a) = [0]
   precedence:
    c > d ~ b ~ a
   problem:
    
   Qed