YES(O(1),O(n^1))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).

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'())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.

Trs:
  { f(g(i(a(), b(), b'()), c()), d()) ->
    if(e(), f(.(b(), c()), d'()), f(.(b'(), c()), d'())) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA).
  
         [f](x1, x2) = [2] x1 + [2] x2 + [0]         
                                                     
         [g](x1, x2) = [1] x1 + [1] x2 + [0]         
                                                     
     [i](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [2]
                                                     
                 [a] = [0]                           
                                                     
                 [b] = [0]                           
                                                     
                [b'] = [0]                           
                                                     
                 [c] = [0]                           
                                                     
                 [d] = [0]                           
                                                     
    [if](x1, x2, x3) = [1] x1 + [1] x2 + [1] x3 + [0]
                                                     
                 [e] = [0]                           
                                                     
         [.](x1, x2) = [1] x1 + [1] x2 + [0]         
                                                     
                [d'] = [0]                           
                                                     
         [h](x1, x2) = [1] x1 + [1] x2 + [0]         
  
  This order satisfies the following ordering constraints:
  
    [f(g(i(a(), b(), b'()), c()), d())] =  [4]                                                         
                                        >  [0]                                                         
                                        =  [if(e(), f(.(b(), c()), d'()), f(.(b'(), c()), d'()))]      
                                                                                                       
          [f(g(h(a(), b()), c()), d())] =  [0]                                                         
                                        >= [0]                                                         
                                        =  [if(e(), f(.(b(), g(h(a(), b()), c())), d()), f(c(), d'()))]
                                                                                                       

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).

Strict Trs:
  { f(g(h(a(), b()), c()), d()) ->
    if(e(), f(.(b(), g(h(a(), b()), c())), d()), f(c(), d'())) }
Weak Trs:
  { f(g(i(a(), b(), b'()), c()), d()) ->
    if(e(), f(.(b(), c()), d'()), f(.(b'(), c()), d'())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs:
  { f(g(h(a(), b()), c()), d()) ->
    if(e(), f(.(b(), g(h(a(), b()), c())), d()), f(c(), d'())) }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(n^1)) . These rules are moved into the corresponding weak
component(s).

Sub-proof:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
         [f](x1, x2) = [2 2] x1 + [1 0] x2 + [0]           
                       [0 0]      [0 1]      [0]           
                                                           
         [g](x1, x2) = [1 0] x1 + [1 0] x2 + [0]           
                       [0 0]      [0 0]      [2]           
                                                           
     [i](x1, x2, x3) = [1 0] x1 + [1 0] x2 + [1 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) = [1 0] x1 + [1 0] x2 + [1 1] x3 + [0]
                       [0 0]      [0 0]      [0 0]      [0]
                                                           
                 [e] = [0]                                 
                       [1]                                 
                                                           
         [.](x1, x2) = [1 0] x1 + [1 0] x2 + [0]           
                       [0 0]      [0 0]      [0]           
                                                           
                [d'] = [0]                                 
                       [3]                                 
                                                           
         [h](x1, x2) = [1 0] x1 + [1 0] x2 + [0]           
                       [0 0]      [0 0]      [0]           
  
  This order satisfies the following ordering constraints:
  
    [f(g(i(a(), b(), b'()), c()), d())] = [4]                                                         
                                          [0]                                                         
                                        > [3]                                                         
                                          [0]                                                         
                                        = [if(e(), f(.(b(), c()), d'()), f(.(b'(), c()), d'()))]      
                                                                                                      
          [f(g(h(a(), b()), c()), d())] = [4]                                                         
                                          [0]                                                         
                                        > [3]                                                         
                                          [0]                                                         
                                        = [if(e(), f(.(b(), g(h(a(), b()), c())), d()), f(c(), d'()))]
                                                                                                      

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

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'())) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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