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


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YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [b](x0) = x0 + 3,
   
   [c](x0) = x0 + 4,
   
   [a](x0) = x0 + 2,
   
   [d](x0) = x0 + 5
  orientation:
   a(a(a(x1))) = x1 + 6 >= x1 + 6 = b(b(x1))
   
   b(b(b(x1))) = x1 + 9 >= x1 + 9 = c(d(x1))
   
   c(x1) = x1 + 4 >= x1 + 4 = a(a(x1))
   
   d(x1) = x1 + 5 >= x1 + 4 = c(x1)
  problem:
   a(a(a(x1))) -> b(b(x1))
   b(b(b(x1))) -> c(d(x1))
   c(x1) -> a(a(x1))
  String Reversal Processor:
   a(a(a(x1))) -> b(b(x1))
   b(b(b(x1))) -> d(c(x1))
   c(x1) -> a(a(x1))
   WPO Processor:
    algebra: Sum
    weight function:
     w0 = 0
     w(c) = 5
     w(b) = w(a) = 2
     w(d) = 0
    status function:
     st(c) = st(d) = st(b) = st(a) = [0]
    precedence:
     c ~ d ~ b ~ a
    problem:
     
    Qed