We consider the following Problem:

  Strict Trs:
    {  f(g(i(a(), b(), b'()), c()), d()) ->
       if(e(), f(.(b(), c()), d'()), f(.(b'(), c()), d'()))
     , f(g(h(a(), b()), c()), d()) ->
       if(e(), f(.(b(), g(h(a(), b()), c())), d()), f(c(), d'()))}
  StartTerms: basic terms
  Strategy: innermost

Certificate: YES(?,O(n^1))

Proof:
  We consider the following Problem:
  
    Strict Trs:
      {  f(g(i(a(), b(), b'()), c()), d()) ->
         if(e(), f(.(b(), c()), d'()), f(.(b'(), c()), d'()))
       , f(g(h(a(), b()), c()), d()) ->
         if(e(), f(.(b(), g(h(a(), b()), c())), d()), f(c(), d'()))}
    StartTerms: basic terms
    Strategy: innermost
  
  Certificate: YES(?,O(n^1))
  
  Proof:
    The weightgap principle applies, where following rules are oriented strictly:
    
    TRS Component:
      {  f(g(i(a(), b(), b'()), c()), d()) ->
         if(e(), f(.(b(), c()), d'()), f(.(b'(), c()), d'()))
       , f(g(h(a(), b()), c()), d()) ->
         if(e(), f(.(b(), g(h(a(), b()), c())), d()), f(c(), d'()))}
    
    Interpretation of nonconstant growth:
    -------------------------------------
      The following argument positions are usable:
        Uargs(f) = {}, Uargs(g) = {}, Uargs(i) = {}, Uargs(if) = {},
        Uargs(.) = {}, Uargs(h) = {}
      We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
      Interpretation Functions:
       f(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
                   [0 0]      [0 0]      [0]
       g(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                   [0 0]      [0 0]      [0]
       i(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                       [0 0]      [0 0]      [0 0]      [0]
       a() = [0]
             [0]
       b() = [0]
             [0]
       b'() = [0]
              [0]
       c() = [0]
             [0]
       d() = [0]
             [0]
       if(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [0]
                        [0 0]      [0 0]      [0 0]      [0]
       e() = [0]
             [0]
       .(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                   [0 0]      [0 0]      [0]
       d'() = [0]
              [0]
       h(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
                   [0 0]      [0 0]      [0]
    
    The strictly oriented rules are moved into the weak component.
    
    We consider the following Problem:
    
      Weak Trs:
        {  f(g(i(a(), b(), b'()), c()), d()) ->
           if(e(), f(.(b(), c()), d'()), f(.(b'(), c()), d'()))
         , f(g(h(a(), b()), c()), d()) ->
           if(e(), f(.(b(), g(h(a(), b()), c())), d()), f(c(), d'()))}
      StartTerms: basic terms
      Strategy: innermost
    
    Certificate: YES(O(1),O(1))
    
    Proof:
      We consider the following Problem:
      
        Weak Trs:
          {  f(g(i(a(), b(), b'()), c()), d()) ->
             if(e(), f(.(b(), c()), d'()), f(.(b'(), c()), d'()))
           , f(g(h(a(), b()), c()), d()) ->
             if(e(), f(.(b(), g(h(a(), b()), c())), d()), f(c(), d'()))}
        StartTerms: basic terms
        Strategy: innermost
      
      Certificate: YES(O(1),O(1))
      
      Proof:
        Empty rules are trivially bounded

Hurray, we answered YES(?,O(n^1))