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(s(x), y)) -> f(c(x, s(y)))
  , g(c(x, s(y))) -> g(c(s(x), 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: { g(c(x, s(y))) -> g(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 0] x1 + [1]           
                  [0 0]      [1]           
                                           
    [c](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                  [0 0]      [0 1]      [0]
                                           
        [s](x1) = [1 0] x1 + [0]           
                  [0 1]      [2]           
                                           
        [g](x1) = [1 1] x1 + [0]           
                  [0 0]      [1]           
  
  This order satisfies the following ordering constraints:
  
    [f(c(s(x), y))] =  [1 0] x + [1 0] y + [1]
                       [0 0]     [0 0]     [1]
                    >= [1 0] x + [1 0] y + [1]
                       [0 0]     [0 0]     [1]
                    =  [f(c(x, s(y)))]        
                                              
    [g(c(x, s(y)))] =  [1 0] x + [1 1] y + [2]
                       [0 0]     [0 0]     [1]
                    >  [1 0] x + [1 1] y + [0]
                       [0 0]     [0 0]     [1]
                    =  [g(c(s(x), 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(s(x), y)) -> f(c(x, s(y))) }
Weak Trs: { g(c(x, s(y))) -> g(c(s(x), 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(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) and not(IDA(1)).
  
        [f](x1) = [1 1] x1 + [0]           
                  [0 0]      [1]           
                                           
    [c](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
                  [0 1]      [0 0]      [0]
                                           
        [s](x1) = [1 0] x1 + [0]           
                  [0 1]      [2]           
                                           
        [g](x1) = [1 0] x1 + [1]           
                  [0 0]      [1]           
  
  This order satisfies the following ordering constraints:
  
    [f(c(s(x), y))] =  [1 1] x + [1 0] y + [2]
                       [0 0]     [0 0]     [1]
                    >  [1 1] x + [1 0] y + [0]
                       [0 0]     [0 0]     [1]
                    =  [f(c(x, s(y)))]        
                                              
    [g(c(x, s(y)))] =  [1 0] x + [1 0] y + [1]
                       [0 0]     [0 0]     [1]
                    >= [1 0] x + [1 0] y + [1]
                       [0 0]     [0 0]     [1]
                    =  [g(c(s(x), 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(s(x), y)) -> f(c(x, s(y)))
  , g(c(x, s(y))) -> g(c(s(x), 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))