MAYBE

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs:
  { eq() -> eq()
  , eq() -> true()
  , eq() -> false()
  , inf(X) -> cons()
  , take(0(), X) -> nil()
  , take(s(), cons()) -> cons()
  , length(cons()) -> s()
  , length(nil()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

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

Trs:
  { eq() -> true()
  , eq() -> false()
  , inf(X) -> cons()
  , take(0(), X) -> nil()
  , take(s(), cons()) -> cons()
  , length(cons()) -> s()
  , length(nil()) -> 0() }

The induced complexity on above rules (modulo remaining rules) is
YES(?,O(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(eq) = {}, safe(true) = {}, safe(false) = {}, safe(inf) = {},
   safe(cons) = {}, safe(take) = {}, safe(0) = {}, safe(nil) = {},
   safe(s) = {}, safe(length) = {}
  
  and precedence
  
   take > inf, length > inf, take ~ length .
  
  Following symbols are considered recursive:
  
   {}
  
  The recursion depth is 0.
  
  For your convenience, here are the satisfied ordering constraints:
  
                   eq() >= eq()   
                                  
                   eq() >  true() 
                                  
                   eq() >  false()
                                  
                inf(X;) >  cons() 
                                  
         take(0(),  X;) >  nil()  
                                  
    take(s(),  cons();) >  cons() 
                                  
        length(cons();) >  s()    
                                  
         length(nil();) >  0()    
                                  

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate MAYBE.

Strict Trs: { eq() -> eq() }
Weak Trs:
  { eq() -> true()
  , eq() -> false()
  , inf(X) -> cons()
  , take(0(), X) -> nil()
  , take(s(), cons()) -> cons()
  , length(cons()) -> s()
  , length(nil()) -> 0() }
Obligation:
  innermost runtime complexity
Answer:
  MAYBE

Empty strict component of the problem is NOT empty.

Arrrr..