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


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YES

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

Proof:
 String Reversal Processor:
  b(a(x1)) -> x1
  c(a(x1)) -> b(a(a(c(c(b(x1))))))
  c(b(x1)) -> x1
  DP Processor:
   DPs:
    c#(a(x1)) -> b#(x1)
    c#(a(x1)) -> c#(b(x1))
    c#(a(x1)) -> c#(c(b(x1)))
    c#(a(x1)) -> b#(a(a(c(c(b(x1))))))
   TRS:
    b(a(x1)) -> x1
    c(a(x1)) -> b(a(a(c(c(b(x1))))))
    c(b(x1)) -> x1
   TDG Processor:
    DPs:
     c#(a(x1)) -> b#(x1)
     c#(a(x1)) -> c#(b(x1))
     c#(a(x1)) -> c#(c(b(x1)))
     c#(a(x1)) -> b#(a(a(c(c(b(x1))))))
    TRS:
     b(a(x1)) -> x1
     c(a(x1)) -> b(a(a(c(c(b(x1))))))
     c(b(x1)) -> x1
    graph:
     c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> b#(a(a(c(c(b(x1))))))
     c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> c#(c(b(x1)))
     c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> c#(b(x1))
     c#(a(x1)) -> c#(c(b(x1))) -> c#(a(x1)) -> b#(x1)
     c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> b#(a(a(c(c(b(x1))))))
     c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> c#(c(b(x1)))
     c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> c#(b(x1))
     c#(a(x1)) -> c#(b(x1)) -> c#(a(x1)) -> b#(x1)
    SCC Processor:
     #sccs: 1
     #rules: 2
     #arcs: 8/16
     DPs:
      c#(a(x1)) -> c#(c(b(x1)))
      c#(a(x1)) -> c#(b(x1))
     TRS:
      b(a(x1)) -> x1
      c(a(x1)) -> b(a(a(c(c(b(x1))))))
      c(b(x1)) -> x1
     Arctic Interpretation Processor:
      dimension: 1
      usable rules:
       b(a(x1)) -> x1
       c(a(x1)) -> b(a(a(c(c(b(x1))))))
       c(b(x1)) -> x1
      interpretation:
       [c#](x0) = x0 + 8,
       
       [c](x0) = 1x0 + 0,
       
       [a](x0) = 1x0 + 10,
       
       [b](x0) = -1x0 + 3
      orientation:
       c#(a(x1)) = 1x1 + 10 >= x1 + 8 = c#(c(b(x1)))
       
       c#(a(x1)) = 1x1 + 10 >= -1x1 + 8 = c#(b(x1))
       
       b(a(x1)) = x1 + 9 >= x1 = x1
       
       c(a(x1)) = 2x1 + 11 >= 2x1 + 10 = b(a(a(c(c(b(x1))))))
       
       c(b(x1)) = x1 + 4 >= x1 = x1
      problem:
       DPs:
        
       TRS:
        b(a(x1)) -> x1
        c(a(x1)) -> b(a(a(c(c(b(x1))))))
        c(b(x1)) -> x1
      Qed