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:
  { prime(0()) -> false()
  , prime(s(0())) -> false()
  , prime(s(s(x))) -> prime1(s(s(x)), s(x))
  , prime1(x, 0()) -> false()
  , prime1(x, s(0())) -> true()
  , prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
  , divp(x, y) -> =(rem(x, y), 0()) }
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:
  { prime(0()) -> false()
  , prime(s(0())) -> false()
  , prime(s(s(x))) -> prime1(s(s(x)), s(x))
  , prime1(x, 0()) -> false()
  , prime1(x, s(0())) -> true()
  , prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
  , divp(x, y) -> =(rem(x, y), 0()) }

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(prime) = {}, safe(0) = {}, safe(false) = {}, safe(s) = {1},
   safe(prime1) = {1}, safe(true) = {}, safe(and) = {1, 2},
   safe(not) = {1}, safe(divp) = {1, 2}, safe(=) = {1, 2},
   safe(rem) = {1, 2}
  
  and precedence
  
   prime > prime1, prime > divp, prime1 > divp .
  
  Following symbols are considered recursive:
  
   {prime1}
  
  The recursion depth is 1.
  
  For your convenience, here are the satisfied ordering constraints:
  
               prime(0();) > false()                                                  
                                                                                      
          prime(s(; 0());) > false()                                                  
                                                                                      
       prime(s(; s(; x));) > prime1(s(; x); s(; s(; x)))                              
                                                                                      
            prime1(0(); x) > false()                                                  
                                                                                      
       prime1(s(; 0()); x) > true()                                                   
                                                                                      
    prime1(s(; s(; y)); x) > and(; not(; divp(; s(; s(; y)),  x)),  prime1(s(; y); x))
                                                                                      
             divp(; x,  y) > =(; rem(; x,  y),  0())                                  
                                                                                      

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:
  { prime(0()) -> false()
  , prime(s(0())) -> false()
  , prime(s(s(x))) -> prime1(s(s(x)), s(x))
  , prime1(x, 0()) -> false()
  , prime1(x, s(0())) -> true()
  , prime1(x, s(s(y))) -> and(not(divp(s(s(y)), x)), prime1(x, s(y)))
  , divp(x, y) -> =(rem(x, y), 0()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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