YES(?,O(n^1))
* Step 1: PolyRank WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval1(A,B) -> eval2(A,B)      [A >= 1]                          (?,1)
          1. eval2(A,B) -> eval2(A,-1 + B) [-1 + A >= 0 && A >= 1 && B >= 1] (?,1)
          2. eval2(A,B) -> eval1(-1 + A,B) [-1 + A >= 0 && A >= 1 && 0 >= B] (?,1)
          3. start(A,B) -> eval1(A,B)      True                              (1,1)
        Signature:
          {(eval1,2);(eval2,2);(start,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{0},3->{0}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(eval1) = x1
          p(eval2) = x1
          p(start) = x1
        
        Following rules are strictly oriented:
        [-1 + A >= 0 && A >= 1 && 0 >= B] ==>                
                               eval2(A,B)   = A              
                                            > -1 + A         
                                            = eval1(-1 + A,B)
        
        
        Following rules are weakly oriented:
                                 [A >= 1] ==>                
                               eval1(A,B)   = A              
                                           >= A              
                                            = eval2(A,B)     
        
        [-1 + A >= 0 && A >= 1 && B >= 1] ==>                
                               eval2(A,B)   = A              
                                           >= A              
                                            = eval2(A,-1 + B)
        
                                     True ==>                
                               start(A,B)   = A              
                                           >= A              
                                            = eval1(A,B)     
        
        
* Step 2: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval1(A,B) -> eval2(A,B)      [A >= 1]                          (?,1)
          1. eval2(A,B) -> eval2(A,-1 + B) [-1 + A >= 0 && A >= 1 && B >= 1] (?,1)
          2. eval2(A,B) -> eval1(-1 + A,B) [-1 + A >= 0 && A >= 1 && 0 >= B] (A,1)
          3. start(A,B) -> eval1(A,B)      True                              (1,1)
        Signature:
          {(eval1,2);(eval2,2);(start,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{0},3->{0}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        We propagate bounds from predecessors.
* Step 3: PolyRank WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval1(A,B) -> eval2(A,B)      [A >= 1]                          (1 + A,1)
          1. eval2(A,B) -> eval2(A,-1 + B) [-1 + A >= 0 && A >= 1 && B >= 1] (?,1)    
          2. eval2(A,B) -> eval1(-1 + A,B) [-1 + A >= 0 && A >= 1 && 0 >= B] (A,1)    
          3. start(A,B) -> eval1(A,B)      True                              (1,1)    
        Signature:
          {(eval1,2);(eval2,2);(start,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{0},3->{0}]
        
    + Applied Processor:
        PolyRank {useFarkas = True, withSizebounds = [], shape = Linear}
    + Details:
        We apply a polynomial interpretation of shape linear:
          p(eval1) = x2
          p(eval2) = x2
          p(start) = x2
        
        Following rules are strictly oriented:
        [-1 + A >= 0 && A >= 1 && B >= 1] ==>                
                               eval2(A,B)   = B              
                                            > -1 + B         
                                            = eval2(A,-1 + B)
        
        
        Following rules are weakly oriented:
                                 [A >= 1] ==>                
                               eval1(A,B)   = B              
                                           >= B              
                                            = eval2(A,B)     
        
        [-1 + A >= 0 && A >= 1 && 0 >= B] ==>                
                               eval2(A,B)   = B              
                                           >= B              
                                            = eval1(-1 + A,B)
        
                                     True ==>                
                               start(A,B)   = B              
                                           >= B              
                                            = eval1(A,B)     
        
        
* Step 4: KnowledgePropagation WORST_CASE(?,O(n^1))
    + Considered Problem:
        Rules:
          0. eval1(A,B) -> eval2(A,B)      [A >= 1]                          (1 + A,1)
          1. eval2(A,B) -> eval2(A,-1 + B) [-1 + A >= 0 && A >= 1 && B >= 1] (B,1)    
          2. eval2(A,B) -> eval1(-1 + A,B) [-1 + A >= 0 && A >= 1 && 0 >= B] (A,1)    
          3. start(A,B) -> eval1(A,B)      True                              (1,1)    
        Signature:
          {(eval1,2);(eval2,2);(start,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{0},3->{0}]
        
    + Applied Processor:
        KnowledgePropagation
    + Details:
        The problem is already solved.

YES(?,O(n^1))