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:
  { admit(x, nil()) -> nil()
  , admit(x, .(u, .(v, .(w(), z)))) ->
    cond(=(sum(x, u, v), w()),
         .(u, .(v, .(w(), admit(carry(x, u, v), z)))))
  , cond(true(), y) -> y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We add the following dependency tuples:

Strict DPs:
  { admit^#(x, nil()) -> c_1()
  , admit^#(x, .(u, .(v, .(w(), z)))) ->
    c_2(cond^#(=(sum(x, u, v), w()),
               .(u, .(v, .(w(), admit(carry(x, u, v), z))))),
        admit^#(carry(x, u, v), z))
  , cond^#(true(), y) -> c_3() }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).

Strict DPs:
  { admit^#(x, nil()) -> c_1()
  , admit^#(x, .(u, .(v, .(w(), z)))) ->
    c_2(cond^#(=(sum(x, u, v), w()),
               .(u, .(v, .(w(), admit(carry(x, u, v), z))))),
        admit^#(carry(x, u, v), z))
  , cond^#(true(), y) -> c_3() }
Weak Trs:
  { admit(x, nil()) -> nil()
  , admit(x, .(u, .(v, .(w(), z)))) ->
    cond(=(sum(x, u, v), w()),
         .(u, .(v, .(w(), admit(carry(x, u, v), z)))))
  , cond(true(), y) -> y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We estimate the number of application of {1,3} by applications of
Pre({1,3}) = {2}. Here rules are labeled as follows:

  DPs:
    { 1: admit^#(x, nil()) -> c_1()
    , 2: admit^#(x, .(u, .(v, .(w(), z)))) ->
         c_2(cond^#(=(sum(x, u, v), w()),
                    .(u, .(v, .(w(), admit(carry(x, u, v), z))))),
             admit^#(carry(x, u, v), z))
    , 3: cond^#(true(), y) -> c_3() }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).

Strict DPs:
  { admit^#(x, .(u, .(v, .(w(), z)))) ->
    c_2(cond^#(=(sum(x, u, v), w()),
               .(u, .(v, .(w(), admit(carry(x, u, v), z))))),
        admit^#(carry(x, u, v), z)) }
Weak DPs:
  { admit^#(x, nil()) -> c_1()
  , cond^#(true(), y) -> c_3() }
Weak Trs:
  { admit(x, nil()) -> nil()
  , admit(x, .(u, .(v, .(w(), z)))) ->
    cond(=(sum(x, u, v), w()),
         .(u, .(v, .(w(), admit(carry(x, u, v), z)))))
  , cond(true(), y) -> y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ admit^#(x, nil()) -> c_1()
, cond^#(true(), y) -> c_3() }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).

Strict DPs:
  { admit^#(x, .(u, .(v, .(w(), z)))) ->
    c_2(cond^#(=(sum(x, u, v), w()),
               .(u, .(v, .(w(), admit(carry(x, u, v), z))))),
        admit^#(carry(x, u, v), z)) }
Weak Trs:
  { admit(x, nil()) -> nil()
  , admit(x, .(u, .(v, .(w(), z)))) ->
    cond(=(sum(x, u, v), w()),
         .(u, .(v, .(w(), admit(carry(x, u, v), z)))))
  , cond(true(), y) -> y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:

  { admit^#(x, .(u, .(v, .(w(), z)))) ->
    c_2(cond^#(=(sum(x, u, v), w()),
               .(u, .(v, .(w(), admit(carry(x, u, v), z))))),
        admit^#(carry(x, u, v), z)) }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).

Strict DPs:
  { admit^#(x, .(u, .(v, .(w(), z)))) ->
    c_1(admit^#(carry(x, u, v), z)) }
Weak Trs:
  { admit(x, nil()) -> nil()
  , admit(x, .(u, .(v, .(w(), z)))) ->
    cond(=(sum(x, u, v), w()),
         .(u, .(v, .(w(), admit(carry(x, u, v), z)))))
  , cond(true(), y) -> y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

No rule is usable, rules are removed from the input problem.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).

Strict DPs:
  { admit^#(x, .(u, .(v, .(w(), z)))) ->
    c_1(admit^#(carry(x, u, v), z)) }
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.

DPs:
  { 1: admit^#(x, .(u, .(v, .(w(), z)))) ->
       c_1(admit^#(carry(x, u, v), z)) }

Sub-proof:
----------
  The input was oriented with the instance of 'Small Polynomial Path
  Order (PS,1-bounded)' as induced by the safe mapping
  
   safe(admit) = {}, safe(nil) = {}, safe(.) = {1, 2}, safe(w) = {},
   safe(cond) = {}, safe(=) = {1, 2}, safe(sum) = {1, 2, 3},
   safe(carry) = {1, 2, 3}, safe(true) = {}, safe(admit^#) = {1},
   safe(c_1) = {}, safe(c_2) = {}, safe(cond^#) = {}, safe(c_3) = {},
   safe(c) = {}, safe(c_1) = {}
  
  and precedence
  
   empty .
  
  Following symbols are considered recursive:
  
   {admit^#}
  
  The recursion depth is 1.
  
  Further, following argument filtering is employed:
  
   pi(admit) = [], pi(nil) = [], pi(.) = [1, 2], pi(w) = [],
   pi(cond) = [], pi(=) = [], pi(sum) = [], pi(carry) = [],
   pi(true) = [], pi(admit^#) = [1, 2], pi(c_1) = [], pi(c_2) = [],
   pi(cond^#) = [], pi(c_3) = [], pi(c) = [], pi(c_1) = [1]
  
  Usable defined function symbols are a subset of:
  
   {admit^#, cond^#}
  
  For your convenience, here are the satisfied ordering constraints:
  
    pi(admit^#(x, .(u, .(v, .(w(), z))))) = admit^#(.(; u,  .(; v,  .(; w(),  z))); x)
                                          > c_1(admit^#(z; carry());)                 
                                          = pi(c_1(admit^#(carry(x, u, v), z)))       
                                                                                      

The strictly oriented rules are moved into the weak component.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

Weak DPs:
  { admit^#(x, .(u, .(v, .(w(), z)))) ->
    c_1(admit^#(carry(x, u, v), z)) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.

{ admit^#(x, .(u, .(v, .(w(), z)))) ->
  c_1(admit^#(carry(x, u, v), z)) }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).

Rules: Empty
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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