YES

Problem:
 and(tt(),X) -> activate(X)
 plus(N,0()) -> N
 plus(N,s(M)) -> s(plus(N,M))
 activate(X) -> X

Proof:
 DP Processor:
  DPs:
   and#(tt(),X) -> activate#(X)
   plus#(N,s(M)) -> plus#(N,M)
  TRS:
   and(tt(),X) -> activate(X)
   plus(N,0()) -> N
   plus(N,s(M)) -> s(plus(N,M))
   activate(X) -> X
  Usable Rule Processor:
   DPs:
    and#(tt(),X) -> activate#(X)
    plus#(N,s(M)) -> plus#(N,M)
   TRS:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [plus#](x0, x1) = [1 1 0 1]x1,
     
     [activate#](x0) = [0],
     
     [and#](x0, x1) = [0 1 0 1]x0,
     
               [0 1 0 0]     [1]
               [0 1 0 1]     [0]
     [s](x0) = [0 0 0 0]x0 + [0]
               [1 0 1 0]     [1],
     
            [0]
            [1]
     [tt] = [0]
            [1]
    orientation:
     and#(tt(),X) = [2] >= [0] = activate#(X)
     
     plus#(N,s(M)) = [1 2 1 1]M + [2] >= [1 1 0 1]M = plus#(N,M)
    problem:
     DPs:
      
     TRS:
      
    Qed