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:
  { plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
  , times(X, s(Y)) -> plus(X, times(Y, X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The weightgap principle applies (using the following nonconstant
growth matrix-interpretation)

The following argument positions are usable:
  Uargs(plus) = {2}

TcT has computed the following matrix interpretation satisfying
not(EDA) and not(IDA(1)).

   [plus](x1, x2) = [1] x2 + [5]         
                                         
  [times](x1, x2) = [1] x1 + [1] x2 + [7]
                                         
          [s](x1) = [1] x1 + [7]         

The following symbols are considered usable

  {plus, times}

The order satisfies the following ordering constraints:

  [plus(plus(X, Y), Z)] = [1] Z + [5]           
                        ? [1] Z + [10]          
                        = [plus(X, plus(Y, Z))] 
                                                
       [times(X, s(Y))] = [1] X + [1] Y + [14]  
                        > [1] X + [1] Y + [12]  
                        = [plus(X, times(Y, X))]
                                                

Further, it can be verified that all rules not oriented are covered by the weightgap condition.

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

Strict Trs: { plus(plus(X, Y), Z) -> plus(X, plus(Y, Z)) }
Weak Trs: { times(X, s(Y)) -> plus(X, times(Y, X)) }
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: { plus(plus(X, Y), Z) -> plus(X, plus(Y, Z)) }

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:
----------
  The following argument positions are usable:
    Uargs(plus) = {2}
  
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
     [plus](x1, x2) = [0 1] x1 + [1 0] x2 + [0]
                      [0 1]      [0 1]      [4]
                                               
    [times](x1, x2) = [1 1] x1 + [1 0] x2 + [1]
                      [1 1]      [1 0]      [1]
                                               
            [s](x1) = [1 1] x1 + [7]           
                      [0 0]      [0]           
  
  The following symbols are considered usable
  
    {plus, times}
  
  The order satisfies the following ordering constraints:
  
    [plus(plus(X, Y), Z)] = [0 1] X + [0 1] Y + [1 0] Z + [4]
                            [0 1]     [0 1]     [0 1]     [8]
                          > [0 1] X + [0 1] Y + [1 0] Z + [0]
                            [0 1]     [0 1]     [0 1]     [8]
                          = [plus(X, plus(Y, Z))]            
                                                             
         [times(X, s(Y))] = [1 1] X + [1 1] Y + [8]          
                            [1 1]     [1 1]     [8]          
                          > [1 1] X + [1 1] Y + [1]          
                            [1 1]     [1 1]     [5]          
                          = [plus(X, times(Y, X))]           
                                                             

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:
  { plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
  , times(X, s(Y)) -> plus(X, times(Y, X)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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