/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:
 f(0()) -> s(0())
 f(s(0())) -> s(0())
 f(s(s(x))) -> f(f(s(x)))

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 0 1]     [0]
   [s](x0) = [0 1 0]x0 + [0]
             [0 0 0]     [1],
   
             [1 1 0]     [0]
   [f](x0) = [0 0 0]x0 + [1]
             [1 1 0]     [0],
   
         [0]
   [0] = [1]
         [0]
  orientation:
            [1]    [0]         
   f(0()) = [1] >= [1] = s(0())
            [1]    [1]         
   
               [1]    [0]         
   f(s(0())) = [1] >= [1] = s(0())
               [1]    [1]         
   
                [1 1 1]    [1]    [1 1 1]    [1]             
   f(s(s(x))) = [0 0 0]x + [1] >= [0 0 0]x + [1] = f(f(s(x)))
                [1 1 1]    [1]    [1 1 1]    [1]             
  problem:
   f(s(s(x))) -> f(f(s(x)))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [s](x0) = x0 + 4,
    
    [f](x0) = x0 + 3
   orientation:
    f(s(s(x))) = x + 11 >= x + 10 = f(f(s(x)))
   problem:
    
   Qed