*** 1 Progress [(O(1),O(n^1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        foldr#3(Cons(x16,x6)) -> step_x_f(rev_l(),x16,foldr#3(x6))
        foldr#3(Nil()) -> fleft_op_e_xs_1()
        main(Cons(x8,x9)) -> step_x_f#1(rev_l(),x8,foldr#3(x9),Nil())
        main(Nil()) -> Nil()
        rev_l#2(x8,x10) -> Cons(x10,x8)
        step_x_f#1(rev_l(),x5,fleft_op_e_xs_1(),x3) -> rev_l#2(x3,x5)
        step_x_f#1(rev_l(),x5,step_x_f(x2,x3,x4),x1) -> step_x_f#1(x2,x3,x4,rev_l#2(x1,x5))
      Weak DP Rules:
        
      Weak TRS Rules:
        
      Signature:
        {foldr#3/1,main/1,rev_l#2/2,step_x_f#1/4} / {Cons/2,Nil/0,fleft_op_e_xs_1/0,rev_l/0,step_x_f/3}
      Obligation:
        Innermost
        basic terms: {foldr#3,main,rev_l#2,step_x_f#1}/{Cons,Nil,fleft_op_e_xs_1,rev_l,step_x_f}
    Applied Processor:
      Bounds {initialAutomaton = minimal, enrichment = match}
    Proof:
      The problem is match-bounded by 3.
      The enriched problem is compatible with follwoing automaton.
        Cons_0(2,2) -> 2
        Cons_1(2,2) -> 1
        Cons_2(2,1) -> 1
        Cons_2(2,2) -> 1
        Cons_3(2,1) -> 1
        Cons_3(2,5) -> 1
        Nil_0() -> 2
        Nil_1() -> 1
        Nil_1() -> 5
        fleft_op_e_xs_1_0() -> 2
        fleft_op_e_xs_1_1() -> 1
        fleft_op_e_xs_1_1() -> 4
        foldr#3_0(2) -> 1
        foldr#3_1(2) -> 4
        main_0(2) -> 1
        rev_l_0() -> 2
        rev_l_1() -> 3
        rev_l#2_0(2,2) -> 1
        rev_l#2_1(1,2) -> 1
        rev_l#2_1(2,2) -> 1
        rev_l#2_2(1,2) -> 1
        rev_l#2_2(5,2) -> 1
        step_x_f_0(2,2,2) -> 2
        step_x_f_1(3,2,4) -> 1
        step_x_f_1(3,2,4) -> 4
        step_x_f#1_0(2,2,2,2) -> 1
        step_x_f#1_1(2,2,2,1) -> 1
        step_x_f#1_1(3,2,4,5) -> 1
        step_x_f#1_2(3,2,4,1) -> 1
*** 1.1 Progress [(O(1),O(1))]  ***
    Considered Problem:
      Strict DP Rules:
        
      Strict TRS Rules:
        
      Weak DP Rules:
        
      Weak TRS Rules:
        foldr#3(Cons(x16,x6)) -> step_x_f(rev_l(),x16,foldr#3(x6))
        foldr#3(Nil()) -> fleft_op_e_xs_1()
        main(Cons(x8,x9)) -> step_x_f#1(rev_l(),x8,foldr#3(x9),Nil())
        main(Nil()) -> Nil()
        rev_l#2(x8,x10) -> Cons(x10,x8)
        step_x_f#1(rev_l(),x5,fleft_op_e_xs_1(),x3) -> rev_l#2(x3,x5)
        step_x_f#1(rev_l(),x5,step_x_f(x2,x3,x4),x1) -> step_x_f#1(x2,x3,x4,rev_l#2(x1,x5))
      Signature:
        {foldr#3/1,main/1,rev_l#2/2,step_x_f#1/4} / {Cons/2,Nil/0,fleft_op_e_xs_1/0,rev_l/0,step_x_f/3}
      Obligation:
        Innermost
        basic terms: {foldr#3,main,rev_l#2,step_x_f#1}/{Cons,Nil,fleft_op_e_xs_1,rev_l,step_x_f}
    Applied Processor:
      EmptyProcessor
    Proof:
      The problem is already closed. The intended complexity is O(1).