YES
* Step 1: FromIts YES
    + Considered Problem:
        Rules:
          0. eval1(A,B,C) -> eval2(A,B,C)      [A >= 1 + B]                       (?,1)
          1. eval2(A,B,C) -> eval1(A,1 + B,C)  [-1 + A + -1*B >= 0 && A >= 1 + C] (?,1)
          2. eval2(A,B,C) -> eval1(A,B,1 + C)  [-1 + A + -1*B >= 0 && A >= 1 + C] (?,1)
          3. eval2(A,B,C) -> eval1(-1 + A,B,C) [-1 + A + -1*B >= 0 && C >= A]     (?,1)
          4. start(A,B,C) -> eval1(A,B,C)      True                               (1,1)
        Signature:
          {(eval1,3);(eval2,3);(start,3)}
        Flow Graph:
          [0->{1,2,3},1->{0},2->{0},3->{0},4->{0}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 2: Decompose YES
    + Considered Problem:
        Rules:
          eval1(A,B,C) -> eval2(A,B,C)       [A >= 1 + B]
          eval2(A,B,C) -> eval1(A,1 + B,C)   [-1 + A + -1*B >= 0 && A >= 1 + C]
          eval2(A,B,C) -> eval1(A,B,1 + C)   [-1 + A + -1*B >= 0 && A >= 1 + C]
          eval2(A,B,C) -> eval1(-1 + A,B,C)  [-1 + A + -1*B >= 0 && C >= A]
          start(A,B,C) -> eval1(A,B,C)       True
        Signature:
          {(eval1,3);(eval2,3);(start,3)}
        Rule Graph:
          [0->{1,2,3},1->{0},2->{0},3->{0},4->{0}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[0,1,2,3] c: [3]
             |
             `- p:[0,1,2] c: [2]
                |
                `- p:[0,1] c: [0,1]
* Step 3: CloseWith YES
    + Considered Problem:
        (Rules:
          eval1(A,B,C) -> eval2(A,B,C)       [A >= 1 + B]
          eval2(A,B,C) -> eval1(A,1 + B,C)   [-1 + A + -1*B >= 0 && A >= 1 + C]
          eval2(A,B,C) -> eval1(A,B,1 + C)   [-1 + A + -1*B >= 0 && A >= 1 + C]
          eval2(A,B,C) -> eval1(-1 + A,B,C)  [-1 + A + -1*B >= 0 && C >= A]
          start(A,B,C) -> eval1(A,B,C)       True
        Signature:
          {(eval1,3);(eval2,3);(start,3)}
        Rule Graph:
          [0->{1,2,3},1->{0},2->{0},3->{0},4->{0}]
        ,We construct a looptree:
          P: [0,1,2,3,4]
          |
          `- p:[0,1,2,3] c: [3]
             |
             `- p:[0,1,2] c: [2]
                |
                `- p:[0,1] c: [0,1])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES