/export/starexec/sandbox/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES

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

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