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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [b](x0) = x0 + 1,
   
   [a](x0) = x0 + 3,
   
   [c](x0) = 3x0 + 1
  orientation:
   a(c(x1)) = 3x1 + 4 >= 3x1 + 4 = c(b(x1))
   
   a(x1) = x1 + 3 >= x1 + 3 = b(b(b(x1)))
   
   b(c(b(x1))) = 3x1 + 5 >= 3x1 + 4 = a(c(x1))
  problem:
   a(c(x1)) -> c(b(x1))
   a(x1) -> b(b(b(x1)))
  KBO Processor:
   weight function:
    w0 = 1
    w(a) = w(c) = 1
    w(b) = 0
   precedence:
    b > a ~ c
   problem:
    
   Qed