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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 1 0]     [0]
   [b](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 1 1]     [0]
   [a](x0) = [0 0 0]x0 + [1]
             [1 0 1]     [1]
  orientation:
                 [1 1 1]     [2]    [1 1 1]     [1]           
   a(b(a(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(a(x1))
                 [1 1 1]     [2]    [0 0 0]     [0]           
   
                 [1 1 0]     [2]    [1 1 0]     [2]              
   b(b(b(x1))) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(a(b(x1)))
                 [0 0 0]     [0]    [0 0 0]     [0]              
   
              [2 1 2]     [2]    [1 1 0]     [2]              
   a(a(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(b(b(x1)))
              [2 1 2]     [2]    [0 0 0]     [0]              
  problem:
   b(b(b(x1))) -> b(a(b(x1)))
   a(a(x1)) -> b(b(b(x1)))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {5,1}
    transitions:
     b1(11) -> 12*
     b1(13) -> 14*
     b1(15) -> 16*
     f20() -> 2*
     b0(6) -> 5*
     b0(2) -> 3*
     b0(4) -> 1*
     b0(3) -> 6*
     a0(3) -> 4*
     a1(12) -> 13*
     14 -> 5*
     16 -> 12*
     4 -> 15*
     2 -> 11*
     1 -> 3,6
   problem:
    
   Qed