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(c(X, s(Y))) -> f(c(s(X), Y))
  , g(c(s(X), Y)) -> f(c(X, s(Y))) }
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: { g(c(s(X), Y)) -> f(c(X, s(Y))) }

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) = [1] x1 + [0]         
                                       
    [c](x1, x2) = [1] x1 + [1] x2 + [0]
                                       
        [s](x1) = [1] x1 + [0]         
                                       
        [g](x1) = [2] x1 + [3]         
  
  This order satisfies the following ordering constraints:
  
    [f(c(X, s(Y)))] =  [1] X + [1] Y + [0]
                    >= [1] X + [1] Y + [0]
                    =  [f(c(s(X), Y))]    
                                          
    [g(c(s(X), Y))] =  [2] X + [2] Y + [3]
                    >  [1] X + [1] Y + [0]
                    =  [f(c(X, s(Y)))]    
                                          

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(c(X, s(Y))) -> f(c(s(X), Y)) }
Weak Trs: { g(c(s(X), Y)) -> f(c(X, s(Y))) }
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(c(X, s(Y))) -> f(c(s(X), Y)) }

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) = [1 1] x1 + [0]           
                  [0 0]      [0]           
                                           
    [c](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                  [0 0]      [0 1]      [0]
                                           
        [s](x1) = [1 0] x1 + [0]           
                  [0 1]      [1]           
                                           
        [g](x1) = [1 1] x1 + [3]           
                  [0 0]      [3]           
  
  This order satisfies the following ordering constraints:
  
    [f(c(X, s(Y)))] = [1 0] X + [1 1] Y + [1]
                      [0 0]     [0 0]     [0]
                    > [1 0] X + [1 1] Y + [0]
                      [0 0]     [0 0]     [0]
                    = [f(c(s(X), Y))]        
                                             
    [g(c(s(X), Y))] = [1 0] X + [1 1] Y + [3]
                      [0 0]     [0 0]     [3]
                    > [1 0] X + [1 1] Y + [1]
                      [0 0]     [0 0]     [0]
                    = [f(c(X, s(Y)))]        
                                             

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(c(X, s(Y))) -> f(c(s(X), Y))
  , g(c(s(X), Y)) -> f(c(X, s(Y))) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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