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

We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.

Trs:
  { and(tt(), X) -> activate(X)
  , activate(X) -> X
  , plus(N, 0()) -> N
  , plus(N, s(M)) -> s(plus(N, M)) }

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 input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(and) = {1}, safe(tt) = {}, safe(activate) = {1},
   safe(plus) = {}, safe(0) = {}, safe(s) = {1}
  
  and precedence
  
   and > activate, plus > activate, and ~ plus .
  
  Following symbols are considered recursive:
  
   {and, plus}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
         and(X; tt()) > activate(; X)    
                                         
        activate(; X) > X                
                                         
       plus(N,  0();) > N                
                                         
    plus(N,  s(; M);) > s(; plus(N,  M;))
                                         

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:
  { and(tt(), X) -> activate(X)
  , activate(X) -> X
  , plus(N, 0()) -> N
  , plus(N, s(M)) -> s(plus(N, M)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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