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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 1 0]     [0]
   [b](x0) = [0 0 1]x0 + [0]
             [0 0 0]     [1],
   
             [1 0 1]  
   [a](x0) = [0 0 1]x0
             [0 0 0]  
  orientation:
                 [1 0 2]     [1]    [1 0 2]     [1]                 
   a(b(a(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = a(b(b(a(x1))))
                 [0 0 0]     [0]    [0 0 0]     [0]                 
   
                 [1 1 1]     [1]    [1 1 1]     [0]           
   b(b(b(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(b(x1))
                 [0 0 0]     [1]    [0 0 0]     [1]           
  problem:
   a(b(a(x1))) -> a(b(b(a(x1))))
  Bounds Processor:
   bound: 0
   enrichment: match
   automaton:
    final states: {1}
    transitions:
     f20() -> 2*
     a0(5) -> 1*
     a0(2) -> 3*
     b0(4) -> 5*
     b0(3) -> 4*
     1 -> 3*
   problem:
    
   Qed