/export/starexec/sandbox2/solver/bin/starexec_run_ttt2-1.17+nonreach /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES

Problem:
 a(p(x1)) -> p(a(A(x1)))
 a(A(x1)) -> A(a(x1))
 p(A(A(x1))) -> a(p(x1))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 1]     [1]
   [A](x0) = [0 0 1]x0 + [0]
             [0 1 0]     [0],
   
             [1 0 1]  
   [a](x0) = [0 0 1]x0
             [0 1 0]  ,
   
             [1 0 0]     [0]
   [p](x0) = [0 1 1]x0 + [1]
             [0 1 1]     [1]
  orientation:
              [1 1 1]     [1]    [1 1 1]     [1]              
   a(p(x1)) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = p(a(A(x1)))
              [0 1 1]     [1]    [0 1 1]     [1]              
   
              [1 1 1]     [1]    [1 1 1]     [1]           
   a(A(x1)) = [0 1 0]x1 + [0] >= [0 1 0]x1 + [0] = A(a(x1))
              [0 0 1]     [0]    [0 0 1]     [0]           
   
                 [1 1 1]     [2]    [1 1 1]     [1]           
   p(A(A(x1))) = [0 1 1]x1 + [1] >= [0 1 1]x1 + [1] = a(p(x1))
                 [0 1 1]     [1]    [0 1 1]     [1]           
  problem:
   a(p(x1)) -> p(a(A(x1)))
   a(A(x1)) -> A(a(x1))
  Bounds Processor:
   bound: 1
   enrichment: match
   automaton:
    final states: {5,1}
    transitions:
     A1(8) -> 9*
     a1(7) -> 8*
     f30() -> 2*
     p0(4) -> 1*
     a0(2) -> 6*
     a0(3) -> 4*
     A0(2) -> 3*
     A0(6) -> 5*
     1 -> 8,6
     2 -> 7*
     5 -> 8,6
     9 -> 4*
   problem:
    
   Qed