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
  TDG 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
   graph:
    plus#(N,s(M)) -> plus#(N,M) -> plus#(N,s(M)) -> plus#(N,M)
   SCC Processor:
    #sccs: 1
    #rules: 1
    #arcs: 1/4
    DPs:
     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
    Subterm Criterion Processor:
     simple projection:
      pi(plus#) = 1
     problem:
      DPs:
       
      TRS:
       and(tt(),X) -> activate(X)
       plus(N,0()) -> N
       plus(N,s(M)) -> s(plus(N,M))
       activate(X) -> X
     Qed