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


--------------------------------------------------------------------------------


YES

Problem:
 a(b(c(x1))) -> c(b(a(x1)))
 C(B(A(x1))) -> A(B(C(x1)))
 b(a(C(x1))) -> C(a(b(x1)))
 c(A(B(x1))) -> B(A(c(x1)))
 A(c(b(x1))) -> b(c(A(x1)))
 B(C(a(x1))) -> a(C(B(x1)))
 a(A(x1)) -> x1
 A(a(x1)) -> x1
 b(B(x1)) -> x1
 B(b(x1)) -> x1
 c(C(x1)) -> x1
 C(c(x1)) -> x1

Proof:
 String Reversal Processor:
  c(b(a(x1))) -> a(b(c(x1)))
  A(B(C(x1))) -> C(B(A(x1)))
  C(a(b(x1))) -> b(a(C(x1)))
  B(A(c(x1))) -> c(A(B(x1)))
  b(c(A(x1))) -> A(c(b(x1)))
  a(C(B(x1))) -> B(C(a(x1)))
  A(a(x1)) -> x1
  a(A(x1)) -> x1
  B(b(x1)) -> x1
  b(B(x1)) -> x1
  C(c(x1)) -> x1
  c(C(x1)) -> x1
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [C](x0) = 2x0 + 1,
    
    [B](x0) = 3x0 + 2,
    
    [A](x0) = x0,
    
    [a](x0) = 5x0 + 4,
    
    [b](x0) = x0,
    
    [c](x0) = x0
   orientation:
    c(b(a(x1))) = 5x1 + 4 >= 5x1 + 4 = a(b(c(x1)))
    
    A(B(C(x1))) = 6x1 + 5 >= 6x1 + 5 = C(B(A(x1)))
    
    C(a(b(x1))) = 10x1 + 9 >= 10x1 + 9 = b(a(C(x1)))
    
    B(A(c(x1))) = 3x1 + 2 >= 3x1 + 2 = c(A(B(x1)))
    
    b(c(A(x1))) = x1 >= x1 = A(c(b(x1)))
    
    a(C(B(x1))) = 30x1 + 29 >= 30x1 + 29 = B(C(a(x1)))
    
    A(a(x1)) = 5x1 + 4 >= x1 = x1
    
    a(A(x1)) = 5x1 + 4 >= x1 = x1
    
    B(b(x1)) = 3x1 + 2 >= x1 = x1
    
    b(B(x1)) = 3x1 + 2 >= x1 = x1
    
    C(c(x1)) = 2x1 + 1 >= x1 = x1
    
    c(C(x1)) = 2x1 + 1 >= x1 = x1
   problem:
    c(b(a(x1))) -> a(b(c(x1)))
    A(B(C(x1))) -> C(B(A(x1)))
    C(a(b(x1))) -> b(a(C(x1)))
    B(A(c(x1))) -> c(A(B(x1)))
    b(c(A(x1))) -> A(c(b(x1)))
    a(C(B(x1))) -> B(C(a(x1)))
   Bounds Processor:
    bound: 0
    enrichment: match
    automaton:
     final states: {17,14,11,8,5,1}
     transitions:
      f60() -> 2*
      a0(2) -> 18*
      a0(9) -> 10*
      a0(4) -> 1*
      b0(10) -> 8*
      b0(2) -> 15*
      b0(3) -> 4*
      c0(15) -> 16*
      c0(2) -> 3*
      c0(13) -> 11*
      C0(7) -> 5*
      C0(2) -> 9*
      C0(18) -> 19*
      B0(2) -> 12*
      B0(19) -> 17*
      B0(6) -> 7*
      A0(12) -> 13*
      A0(2) -> 6*
      A0(16) -> 14*
      1 -> 3,16
      5 -> 6,13
      8 -> 9,19
      11 -> 12,7
      14 -> 15,4
      17 -> 18,10
    problem:
     
    Qed