*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        activate(X) -> X
        activate(n__f(X)) -> f(X)
        f(X) -> if(X,c(),n__f(true()))
        f(X) -> n__f(X)
        if(false(),X,Y) -> activate(Y)
        if(true(),X,Y) -> X
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {activate/1,f/1,if/3} / {c/0,false/0,n__f/1,true/0}
      Obligation:
        Innermost
        basic terms: {activate,f,if}/{c,false,n__f,true}
    Applied Processor:
      Bounds {initialAutomaton = minimal, enrichment = match}
    Proof:
      The problem is match-bounded by 3.
      The enriched problem is compatible with follwoing automaton.
        activate_0(2) -> 1
        activate_1(2) -> 1
        activate_1(4) -> 1
        activate_1(7) -> 1
        c_0() -> 1
        c_0() -> 2
        c_1() -> 1
        c_1() -> 3
        c_2() -> 1
        c_2() -> 6
        c_3() -> 1
        c_3() -> 9
        f_0(2) -> 1
        f_1(2) -> 1
        f_2(5) -> 1
        f_2(8) -> 1
        false_0() -> 1
        false_0() -> 2
        if_0(2,2,2) -> 1
        if_1(2,3,4) -> 1
        if_2(2,6,7) -> 1
        if_3(5,9,10) -> 1
        if_3(8,9,10) -> 1
        n__f_0(2) -> 1
        n__f_0(2) -> 2
        n__f_1(2) -> 1
        n__f_1(5) -> 1
        n__f_1(5) -> 4
        n__f_2(2) -> 1
        n__f_2(8) -> 1
        n__f_2(8) -> 7
        n__f_3(5) -> 1
        n__f_3(8) -> 1
        n__f_3(11) -> 10
        true_0() -> 1
        true_0() -> 2
        true_1() -> 5
        true_2() -> 8
        true_3() -> 11
        2 -> 1
        3 -> 1
        4 -> 1
        6 -> 1
        7 -> 1
        9 -> 1
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        activate(X) -> X
        activate(n__f(X)) -> f(X)
        f(X) -> if(X,c(),n__f(true()))
        f(X) -> n__f(X)
        if(false(),X,Y) -> activate(Y)
        if(true(),X,Y) -> X
      Signature:
        {activate/1,f/1,if/3} / {c/0,false/0,n__f/1,true/0}
      Obligation:
        Innermost
        basic terms: {activate,f,if}/{c,false,n__f,true}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).