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:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

We add the following weak dependency pairs:

Strict DPs:
  { plus^#(plus(X, Y), Z) -> c_1(plus^#(X, plus(Y, Z)))
  , times^#(X, s(Y)) -> c_2(plus^#(X, times(Y, X))) }

and mark the set of starting terms.

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

Strict DPs:
  { plus^#(plus(X, Y), Z) -> c_1(plus^#(X, plus(Y, Z)))
  , times^#(X, s(Y)) -> c_2(plus^#(X, times(Y, X))) }
Strict Trs:
  { plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
  , times(X, s(Y)) -> plus(X, times(Y, X)) }
Obligation:
  runtime complexity
Answer:
  YES(O(1),O(n^1))

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

The following argument positions are usable:
  Uargs(plus) = {2}, Uargs(plus^#) = {2}, Uargs(c_1) = {1},
  Uargs(c_2) = {1}

TcT has computed the following constructor-restricted matrix
interpretation.

     [plus](x1, x2) = [0 1] x1 + [1 0] x2 + [0]
                      [0 1]      [0 1]      [1]
                                               
    [times](x1, x2) = [1 2] x1 + [1 0] x2 + [0]
                      [2 1]      [2 0]      [0]
                                               
            [s](x1) = [1 2] x1 + [2]           
                      [0 0]      [0]           
                                               
   [plus^#](x1, x2) = [0 1] x1 + [1 0] x2 + [0]
                      [0 0]      [0 0]      [0]
                                               
          [c_1](x1) = [1 0] x1 + [0]           
                      [0 1]      [0]           
                                               
  [times^#](x1, x2) = [2 1] x1 + [1 0] x2 + [0]
                      [0 0]      [0 0]      [0]
                                               
          [c_2](x1) = [1 0] x1 + [0]           
                      [0 1]      [0]           

The order satisfies the following ordering constraints:

    [plus(plus(X, Y), Z)] = [0 1] X + [0 1] Y + [1 0] Z + [1]
                            [0 1]     [0 1]     [0 1]     [2]
                          > [0 1] X + [0 1] Y + [1 0] Z + [0]
                            [0 1]     [0 1]     [0 1]     [2]
                          = [plus(X, plus(Y, Z))]            
                                                             
         [times(X, s(Y))] = [1 2] X + [1 2] Y + [2]          
                            [2 1]     [2 4]     [4]          
                          > [1 1] X + [1 2] Y + [0]          
                            [2 1]     [2 1]     [1]          
                          = [plus(X, times(Y, X))]           
                                                             
  [plus^#(plus(X, Y), Z)] = [0 1] X + [0 1] Y + [1 0] Z + [1]
                            [0 0]     [0 0]     [0 0]     [0]
                          > [0 1] X + [0 1] Y + [1 0] Z + [0]
                            [0 0]     [0 0]     [0 0]     [0]
                          = [c_1(plus^#(X, plus(Y, Z)))]     
                                                             
       [times^#(X, s(Y))] = [2 1] X + [1 2] Y + [2]          
                            [0 0]     [0 0]     [0]          
                          > [1 1] X + [1 2] Y + [0]          
                            [0 0]     [0 0]     [0]          
                          = [c_2(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(n^1)).

Weak DPs:
  { plus^#(plus(X, Y), Z) -> c_1(plus^#(X, plus(Y, Z)))
  , times^#(X, s(Y)) -> c_2(plus^#(X, times(Y, X))) }
Weak Trs:
  { plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
  , times(X, s(Y)) -> plus(X, times(Y, X)) }
Obligation:
  runtime complexity
Answer:
  YES(?,O(n^1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ plus^#(plus(X, Y), Z) -> c_1(plus^#(X, plus(Y, Z)))
, times^#(X, s(Y)) -> c_2(plus^#(X, times(Y, X))) }

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

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

We employ 'linear path analysis' using the following approximated
dependency graph:
empty


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