YES
* Step 1: UnsatPaths YES
    + Considered Problem:
        Rules:
          0. f1(A,B) -> f3(A,A)           [A >= 1]                                           (1,1)
          1. f3(A,B) -> f3(A,-1 + B)      [A + -1*B >= 0 && -1 + A >= 0 && B >= 1]           (?,1)
          2. f3(A,B) -> f3(-1 + A,-1 + A) [A + -1*B >= 0 && -1 + A >= 0 && 0 >= B && A >= 2] (?,1)
        Signature:
          {(f1,2);(f3,2)}
        Flow Graph:
          [0->{1,2},1->{1,2},2->{1,2}]
        
    + Applied Processor:
        UnsatPaths
    + Details:
        We remove following edges from the transition graph: [(0,2),(2,2)]
* Step 2: FromIts YES
    + Considered Problem:
        Rules:
          0. f1(A,B) -> f3(A,A)           [A >= 1]                                           (1,1)
          1. f3(A,B) -> f3(A,-1 + B)      [A + -1*B >= 0 && -1 + A >= 0 && B >= 1]           (?,1)
          2. f3(A,B) -> f3(-1 + A,-1 + A) [A + -1*B >= 0 && -1 + A >= 0 && 0 >= B && A >= 2] (?,1)
        Signature:
          {(f1,2);(f3,2)}
        Flow Graph:
          [0->{1},1->{1,2},2->{1}]
        
    + Applied Processor:
        FromIts
    + Details:
        ()
* Step 3: Decompose YES
    + Considered Problem:
        Rules:
          f1(A,B) -> f3(A,A)            [A >= 1]
          f3(A,B) -> f3(A,-1 + B)       [A + -1*B >= 0 && -1 + A >= 0 && B >= 1]
          f3(A,B) -> f3(-1 + A,-1 + A)  [A + -1*B >= 0 && -1 + A >= 0 && 0 >= B && A >= 2]
        Signature:
          {(f1,2);(f3,2)}
        Rule Graph:
          [0->{1},1->{1,2},2->{1}]
    + Applied Processor:
        Decompose NoGreedy
    + Details:
        We construct a looptree:
          P: [0,1,2]
          |
          `- p:[1,2] c: [2]
             |
             `- p:[1] c: [1]
* Step 4: CloseWith YES
    + Considered Problem:
        (Rules:
          f1(A,B) -> f3(A,A)            [A >= 1]
          f3(A,B) -> f3(A,-1 + B)       [A + -1*B >= 0 && -1 + A >= 0 && B >= 1]
          f3(A,B) -> f3(-1 + A,-1 + A)  [A + -1*B >= 0 && -1 + A >= 0 && 0 >= B && A >= 2]
        Signature:
          {(f1,2);(f3,2)}
        Rule Graph:
          [0->{1},1->{1,2},2->{1}]
        ,We construct a looptree:
          P: [0,1,2]
          |
          `- p:[1,2] c: [2]
             |
             `- p:[1] c: [1])
    + Applied Processor:
        CloseWith True
    + Details:
        ()

YES