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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = 4x0,
   
   [b](x0) = x0 + 1,
   
   [a](x0) = x0 + 4,
   
   [d](x0) = 4x0
  orientation:
   a(d(x1)) = 4x1 + 4 >= 4x1 + 4 = d(b(x1))
   
   a(x1) = x1 + 4 >= x1 + 3 = b(b(b(x1)))
   
   b(d(b(x1))) = 4x1 + 5 >= 4x1 + 4 = a(c(x1))
   
   c(x1) = 4x1 >= 4x1 = d(x1)
  problem:
   a(d(x1)) -> d(b(x1))
   c(x1) -> d(x1)
  String Reversal Processor:
   d(a(x1)) -> b(d(x1))
   c(x1) -> d(x1)
   KBO Processor:
    weight function:
     w0 = 1
     w(c) = w(b) = w(a) = w(d) = 1
    precedence:
     a > c > d > b
    problem:
     
    Qed