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:
  { U11(tt(), M, N) -> U12(tt(), M, N)
  , U12(tt(), M, N) -> s(plus(N, M))
  , plus(N, s(M)) -> U11(tt(), M, N)
  , plus(N, 0()) -> N }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add following weak dependency pairs:

Strict DPs:
  { U11^#(tt(), M, N) -> c_1(U12^#(tt(), M, N))
  , U12^#(tt(), M, N) -> c_2(plus^#(N, M))
  , plus^#(N, s(M)) -> c_3(U11^#(tt(), M, N))
  , plus^#(N, 0()) -> c_4() }

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:
  { U11^#(tt(), M, N) -> c_1(U12^#(tt(), M, N))
  , U12^#(tt(), M, N) -> c_2(plus^#(N, M))
  , plus^#(N, s(M)) -> c_3(U11^#(tt(), M, N))
  , plus^#(N, 0()) -> c_4() }
Strict Trs:
  { U11(tt(), M, N) -> U12(tt(), M, N)
  , U12(tt(), M, N) -> s(plus(N, M))
  , plus(N, s(M)) -> U11(tt(), M, N)
  , plus(N, 0()) -> N }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

No rule is usable, rules are removed from the input problem.

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

Strict DPs:
  { U11^#(tt(), M, N) -> c_1(U12^#(tt(), M, N))
  , U12^#(tt(), M, N) -> c_2(plus^#(N, M))
  , plus^#(N, s(M)) -> c_3(U11^#(tt(), M, N))
  , plus^#(N, 0()) -> c_4() }
Obligation:
  innermost 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(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1}

TcT has computed following constructor-restricted matrix
interpretation.

                 [tt] = [0]         
                                    
              [s](x1) = [1] x1 + [2]
                                    
                  [0] = [1]         
                                    
  [U11^#](x1, x2, x3) = [2] x3 + [0]
                                    
            [c_1](x1) = [1] x1 + [1]
                                    
  [U12^#](x1, x2, x3) = [2] x3 + [0]
                                    
            [c_2](x1) = [1] x1 + [1]
                                    
     [plus^#](x1, x2) = [2] x1 + [1]
                                    
            [c_3](x1) = [1] x1 + [1]
                                    
                [c_4] = [0]         

This order satisfies following ordering constraints:


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)).

Strict DPs:
  { U11^#(tt(), M, N) -> c_1(U12^#(tt(), M, N))
  , U12^#(tt(), M, N) -> c_2(plus^#(N, M))
  , plus^#(N, s(M)) -> c_3(U11^#(tt(), M, N)) }
Weak DPs: { plus^#(N, 0()) -> c_4() }
Obligation:
  innermost 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^#(N, 0()) -> c_4() }

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

Strict DPs:
  { U11^#(tt(), M, N) -> c_1(U12^#(tt(), M, N))
  , U12^#(tt(), M, N) -> c_2(plus^#(N, M))
  , plus^#(N, s(M)) -> c_3(U11^#(tt(), M, N)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(?,O(n^1))

The following argument positions are usable:
  Uargs(c_1) = {1}, Uargs(c_2) = {1}, Uargs(c_3) = {1}

TcT has computed following constructor-based matrix interpretation
satisfying not(EDA).

                 [tt] = [0]         
                                    
              [s](x1) = [1] x1 + [2]
                                    
  [U11^#](x1, x2, x3) = [2] x2 + [3]
                                    
            [c_1](x1) = [1] x1 + [1]
                                    
  [U12^#](x1, x2, x3) = [2] x2 + [1]
                                    
            [c_2](x1) = [1] x1 + [0]
                                    
     [plus^#](x1, x2) = [2] x2 + [0]
                                    
            [c_3](x1) = [1] x1 + [0]

This order satisfies following ordering constraints:

  [U11^#(tt(), M, N)] = [2] M + [3]             
                      > [2] M + [2]             
                      = [c_1(U12^#(tt(), M, N))]
                                                
  [U12^#(tt(), M, N)] = [2] M + [1]             
                      > [2] M + [0]             
                      = [c_2(plus^#(N, M))]     
                                                
    [plus^#(N, s(M))] = [2] M + [4]             
                      > [2] M + [3]             
                      = [c_3(U11^#(tt(), M, N))]
                                                

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