*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        activate(X) -> X
        activate(n__f(X)) -> f(activate(X))
        activate(n__true()) -> true()
        f(X) -> if(X,c(),n__f(n__true()))
        f(X) -> n__f(X)
        if(false(),X,Y) -> activate(Y)
        if(true(),X,Y) -> X
        true() -> n__true()
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {activate/1,f/1,if/3,true/0} / {c/0,false/0,n__f/1,n__true/0}
      Obligation:
        Innermost
        basic terms: {activate,f,if,true}/{c,false,n__f,n__true}
    Applied Processor:
      InnermostRuleRemoval
    Proof:
      Arguments of following rules are not normal-forms.
        if(true(),X,Y) -> X
      All above mentioned rules can be savely removed.
*** 1.1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        activate(X) -> X
        activate(n__f(X)) -> f(activate(X))
        activate(n__true()) -> true()
        f(X) -> if(X,c(),n__f(n__true()))
        f(X) -> n__f(X)
        if(false(),X,Y) -> activate(Y)
        true() -> n__true()
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {activate/1,f/1,if/3,true/0} / {c/0,false/0,n__f/1,n__true/0}
      Obligation:
        Innermost
        basic terms: {activate,f,if,true}/{c,false,n__f,n__true}
    Applied Processor:
      Bounds {initialAutomaton = minimal, enrichment = match}
    Proof:
      The problem is match-bounded by 4.
      The enriched problem is compatible with follwoing automaton.
        activate_0(2) -> 1
        activate_1(2) -> 1
        activate_1(2) -> 3
        activate_1(5) -> 1
        activate_1(8) -> 1
        activate_1(8) -> 3
        activate_2(6) -> 10
        activate_2(9) -> 10
        c_0() -> 1
        c_0() -> 2
        c_0() -> 3
        c_1() -> 4
        c_2() -> 7
        c_3() -> 11
        f_0(2) -> 1
        f_1(3) -> 1
        f_1(3) -> 3
        f_2(10) -> 1
        f_2(10) -> 3
        false_0() -> 1
        false_0() -> 2
        false_0() -> 3
        if_0(2,2,2) -> 1
        if_1(2,4,5) -> 1
        if_2(3,7,8) -> 1
        if_2(3,7,8) -> 3
        if_3(10,11,12) -> 1
        if_3(10,11,12) -> 3
        n__f_0(2) -> 1
        n__f_0(2) -> 2
        n__f_0(2) -> 3
        n__f_1(2) -> 1
        n__f_1(6) -> 1
        n__f_1(6) -> 5
        n__f_2(3) -> 1
        n__f_2(3) -> 3
        n__f_2(9) -> 1
        n__f_2(9) -> 3
        n__f_2(9) -> 8
        n__f_3(10) -> 1
        n__f_3(10) -> 3
        n__f_3(13) -> 12
        n__true_0() -> 1
        n__true_0() -> 2
        n__true_0() -> 3
        n__true_1() -> 1
        n__true_1() -> 6
        n__true_1() -> 10
        n__true_2() -> 1
        n__true_2() -> 3
        n__true_2() -> 9
        n__true_2() -> 10
        n__true_3() -> 10
        n__true_3() -> 13
        n__true_4() -> 10
        true_0() -> 1
        true_1() -> 1
        true_1() -> 3
        true_2() -> 10
        true_3() -> 10
        2 -> 1
        2 -> 3
        5 -> 1
        6 -> 10
        8 -> 1
        8 -> 3
        9 -> 10
*** 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(activate(X))
        activate(n__true()) -> true()
        f(X) -> if(X,c(),n__f(n__true()))
        f(X) -> n__f(X)
        if(false(),X,Y) -> activate(Y)
        true() -> n__true()
      Signature:
        {activate/1,f/1,if/3,true/0} / {c/0,false/0,n__f/1,n__true/0}
      Obligation:
        Innermost
        basic terms: {activate,f,if,true}/{c,false,n__f,n__true}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).