YES(O(1),O(n^3))

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict Trs:
  { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
  , mul0(Z(), y) -> Z()
  , add0(C(x, y), y') -> add0(y, C(S(), y'))
  , add0(Z(), y) -> y
  , second(C(x, y)) -> y
  , isZero(C(x, y)) -> False()
  , isZero(Z()) -> True()
  , goal(xs, ys) -> mul0(xs, ys) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We add the following dependency tuples:

Strict DPs:
  { mul0^#(C(x, y), y') ->
    c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y'))
  , mul0^#(Z(), y) -> c_2()
  , add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y')))
  , add0^#(Z(), y) -> c_4()
  , second^#(C(x, y)) -> c_5()
  , isZero^#(C(x, y)) -> c_6()
  , isZero^#(Z()) -> c_7()
  , goal^#(xs, ys) -> c_8(mul0^#(xs, ys)) }

and mark the set of starting terms.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { mul0^#(C(x, y), y') ->
    c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y'))
  , mul0^#(Z(), y) -> c_2()
  , add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y')))
  , add0^#(Z(), y) -> c_4()
  , second^#(C(x, y)) -> c_5()
  , isZero^#(C(x, y)) -> c_6()
  , isZero^#(Z()) -> c_7()
  , goal^#(xs, ys) -> c_8(mul0^#(xs, ys)) }
Weak Trs:
  { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
  , mul0(Z(), y) -> Z()
  , add0(C(x, y), y') -> add0(y, C(S(), y'))
  , add0(Z(), y) -> y
  , second(C(x, y)) -> y
  , isZero(C(x, y)) -> False()
  , isZero(Z()) -> True()
  , goal(xs, ys) -> mul0(xs, ys) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We estimate the number of application of {2,4,5,6,7} by
applications of Pre({2,4,5,6,7}) = {1,3,8}. Here rules are labeled
as follows:

  DPs:
    { 1: mul0^#(C(x, y), y') ->
         c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y'))
    , 2: mul0^#(Z(), y) -> c_2()
    , 3: add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y')))
    , 4: add0^#(Z(), y) -> c_4()
    , 5: second^#(C(x, y)) -> c_5()
    , 6: isZero^#(C(x, y)) -> c_6()
    , 7: isZero^#(Z()) -> c_7()
    , 8: goal^#(xs, ys) -> c_8(mul0^#(xs, ys)) }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { mul0^#(C(x, y), y') ->
    c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y'))
  , add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y')))
  , goal^#(xs, ys) -> c_8(mul0^#(xs, ys)) }
Weak DPs:
  { mul0^#(Z(), y) -> c_2()
  , add0^#(Z(), y) -> c_4()
  , second^#(C(x, y)) -> c_5()
  , isZero^#(C(x, y)) -> c_6()
  , isZero^#(Z()) -> c_7() }
Weak Trs:
  { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
  , mul0(Z(), y) -> Z()
  , add0(C(x, y), y') -> add0(y, C(S(), y'))
  , add0(Z(), y) -> y
  , second(C(x, y)) -> y
  , isZero(C(x, y)) -> False()
  , isZero(Z()) -> True()
  , goal(xs, ys) -> mul0(xs, ys) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

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

{ mul0^#(Z(), y) -> c_2()
, add0^#(Z(), y) -> c_4()
, second^#(C(x, y)) -> c_5()
, isZero^#(C(x, y)) -> c_6()
, isZero^#(Z()) -> c_7() }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { mul0^#(C(x, y), y') ->
    c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y'))
  , add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y')))
  , goal^#(xs, ys) -> c_8(mul0^#(xs, ys)) }
Weak Trs:
  { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
  , mul0(Z(), y) -> Z()
  , add0(C(x, y), y') -> add0(y, C(S(), y'))
  , add0(Z(), y) -> y
  , second(C(x, y)) -> y
  , isZero(C(x, y)) -> False()
  , isZero(Z()) -> True()
  , goal(xs, ys) -> mul0(xs, ys) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

Consider the dependency graph

  1: mul0^#(C(x, y), y') ->
     c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y'))
     -->_1 add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y'))) :2
     -->_2 mul0^#(C(x, y), y') ->
           c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y')) :1
  
  2: add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y')))
     -->_1 add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y'))) :2
  
  3: goal^#(xs, ys) -> c_8(mul0^#(xs, ys))
     -->_1 mul0^#(C(x, y), y') ->
           c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y')) :1
  

Following roots of the dependency graph are removed, as the
considered set of starting terms is closed under reduction with
respect to these rules (modulo compound contexts).

  { goal^#(xs, ys) -> c_8(mul0^#(xs, ys)) }


We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { mul0^#(C(x, y), y') ->
    c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y'))
  , add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y'))) }
Weak Trs:
  { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
  , mul0(Z(), y) -> Z()
  , add0(C(x, y), y') -> add0(y, C(S(), y'))
  , add0(Z(), y) -> y
  , second(C(x, y)) -> y
  , isZero(C(x, y)) -> False()
  , isZero(Z()) -> True()
  , goal(xs, ys) -> mul0(xs, ys) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We replace rewrite rules by usable rules:

  Weak Usable Rules:
    { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
    , mul0(Z(), y) -> Z()
    , add0(C(x, y), y') -> add0(y, C(S(), y'))
    , add0(Z(), y) -> y }

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^3)).

Strict DPs:
  { mul0^#(C(x, y), y') ->
    c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y'))
  , add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y'))) }
Weak Trs:
  { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
  , mul0(Z(), y) -> Z()
  , add0(C(x, y), y') -> add0(y, C(S(), y'))
  , add0(Z(), y) -> y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^3))

We decompose the input problem according to the dependency graph
into the upper component

  { mul0^#(C(x, y), y') ->
    c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y')) }

and lower component

  { add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y'))) }

Further, following extension rules are added to the lower
component.

{ mul0^#(C(x, y), y') -> mul0^#(y, y')
, mul0^#(C(x, y), y') -> add0^#(mul0(y, y'), y') }

TcT solves the upper component with certificate YES(O(1),O(n^1)).

Sub-proof:
----------
  We are left with following problem, upon which TcT provides the
  certificate YES(O(1),O(n^1)).
  
  Strict DPs:
    { mul0^#(C(x, y), y') ->
      c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y')) }
  Weak Trs:
    { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
    , mul0(Z(), y) -> Z()
    , add0(C(x, y), y') -> add0(y, C(S(), y'))
    , add0(Z(), y) -> y }
  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: mul0^#(C(x, y), y') ->
         c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y')) }
  
  Sub-proof:
  ----------
    The input was oriented with the instance of 'Small Polynomial Path
    Order (PS,1-bounded)' as induced by the safe mapping
    
     safe(mul0) = {}, safe(C) = {1, 2}, safe(add0) = {}, safe(Z) = {},
     safe(S) = {}, safe(second) = {}, safe(isZero) = {},
     safe(False) = {}, safe(True) = {}, safe(goal) = {},
     safe(mul0^#) = {2}, safe(c_1) = {}, safe(add0^#) = {},
     safe(c_2) = {}, safe(c_3) = {}, safe(c_4) = {},
     safe(second^#) = {}, safe(c_5) = {}, safe(isZero^#) = {},
     safe(c_6) = {}, safe(c_7) = {}, safe(goal^#) = {}, safe(c_8) = {}
    
    and precedence
    
     mul0^# > add0, mul0 ~ add0 .
    
    Following symbols are considered recursive:
    
     {mul0^#}
    
    The recursion depth is 1.
    
    Further, following argument filtering is employed:
    
     pi(mul0) = 2, pi(C) = [2], pi(add0) = 2, pi(Z) = [], pi(S) = [],
     pi(second) = [], pi(isZero) = [], pi(False) = [], pi(True) = [],
     pi(goal) = [], pi(mul0^#) = [1], pi(c_1) = [1, 2], pi(add0^#) = [],
     pi(c_2) = [], pi(c_3) = [], pi(c_4) = [], pi(second^#) = [],
     pi(c_5) = [], pi(isZero^#) = [], pi(c_6) = [], pi(c_7) = [],
     pi(goal^#) = [], pi(c_8) = []
    
    Usable defined function symbols are a subset of:
    
     {mul0^#, add0^#, second^#, isZero^#, goal^#}
    
    For your convenience, here are the satisfied ordering constraints:
    
      pi(mul0^#(C(x, y), y')) = mul0^#(C(; y);)                                
                              > c_1(add0^#(),  mul0^#(y;);)                    
                              = pi(c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y')))
                                                                               
  
  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:
    { mul0^#(C(x, y), y') ->
      c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y')) }
  Weak Trs:
    { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
    , mul0(Z(), y) -> Z()
    , add0(C(x, y), y') -> add0(y, C(S(), y'))
    , add0(Z(), y) -> y }
  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.
  
  { mul0^#(C(x, y), y') ->
    c_1(add0^#(mul0(y, y'), y'), mul0^#(y, y')) }
  
  We are left with following problem, upon which TcT provides the
  certificate YES(O(1),O(1)).
  
  Weak Trs:
    { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
    , mul0(Z(), y) -> Z()
    , add0(C(x, y), y') -> add0(y, C(S(), y'))
    , add0(Z(), y) -> y }
  Obligation:
    innermost runtime complexity
  Answer:
    YES(O(1),O(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(1)).
  
  Rules: Empty
  Obligation:
    innermost runtime complexity
  Answer:
    YES(O(1),O(1))
  
  Empty rules are trivially bounded

We return to the main proof.

We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).

Strict DPs: { add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y'))) }
Weak DPs:
  { mul0^#(C(x, y), y') -> mul0^#(y, y')
  , mul0^#(C(x, y), y') -> add0^#(mul0(y, y'), y') }
Weak Trs:
  { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
  , mul0(Z(), y) -> Z()
  , add0(C(x, y), y') -> add0(y, C(S(), y'))
  , add0(Z(), y) -> y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^2))

We use the processor 'polynomial interpretation' to orient
following rules strictly.

DPs:
  { 1: add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y')))
  , 3: mul0^#(C(x, y), y') -> add0^#(mul0(y, y'), y') }

Sub-proof:
----------
  The following argument positions are considered usable:
  
    Uargs(c_3) = {1}
  
  TcT has computed the following constructor-restricted polynomial
  interpretation.
  
    [mul0](x1, x2) = x1*x2                                  
                                                            
       [C](x1, x2) = 1 + x2                                 
                                                            
    [add0](x1, x2) = x1 + x2                                
                                                            
             [Z]() = 0                                      
                                                            
             [S]() = 0                                      
                                                            
      [second](x1) = 3*x1 + 3*x1^2                          
                                                            
      [isZero](x1) = 3*x1 + 3*x1^2                          
                                                            
         [False]() = 0                                      
                                                            
          [True]() = 0                                      
                                                            
    [goal](x1, x2) = 3*x1 + 3*x1*x2 + 3*x1^2 + 3*x2 + 3*x2^2
                                                            
  [mul0^#](x1, x2) = 3 + x1*x2 + 2*x2 + 3*x2^2              
                                                            
     [c_1](x1, x2) = 3*x1 + 3*x1*x2 + 3*x1^2 + 3*x2 + 3*x2^2
                                                            
  [add0^#](x1, x2) = x1                                     
                                                            
           [c_2]() = 0                                      
                                                            
         [c_3](x1) = x1                                     
                                                            
           [c_4]() = 0                                      
                                                            
    [second^#](x1) = 3*x1 + 3*x1^2                          
                                                            
           [c_5]() = 0                                      
                                                            
    [isZero^#](x1) = 3*x1 + 3*x1^2                          
                                                            
           [c_6]() = 0                                      
                                                            
           [c_7]() = 0                                      
                                                            
  [goal^#](x1, x2) = 3*x1 + 3*x1*x2 + 3*x1^2 + 3*x2 + 3*x2^2
                                                            
         [c_8](x1) = 3*x1 + 3*x1^2                          
                                                            
  
  The following symbols are considered usable
  
    {mul0, add0, mul0^#, add0^#, second^#, isZero^#, goal^#}
  
  This order satisfies the following ordering constraints.
  
      [mul0(C(x, y), y')] =  y' + y*y'                   
                          >= y*y' + y'                   
                          =  [add0(mul0(y, y'), y')]     
                                                         
           [mul0(Z(), y)] =                              
                          >=                             
                          =  [Z()]                       
                                                         
      [add0(C(x, y), y')] =  1 + y + y'                  
                          >= y + 1 + y'                  
                          =  [add0(y, C(S(), y'))]       
                                                         
           [add0(Z(), y)] =  y                           
                          >= y                           
                          =  [y]                         
                                                         
    [mul0^#(C(x, y), y')] =  3 + 3*y' + y*y' + 3*y'^2    
                          >= 3 + y*y' + 2*y' + 3*y'^2    
                          =  [mul0^#(y, y')]             
                                                         
    [mul0^#(C(x, y), y')] =  3 + 3*y' + y*y' + 3*y'^2    
                          >  y*y'                        
                          =  [add0^#(mul0(y, y'), y')]   
                                                         
    [add0^#(C(x, y), y')] =  1 + y                       
                          >  y                           
                          =  [c_3(add0^#(y, C(S(), y')))]
                                                         

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:
  { mul0^#(C(x, y), y') -> mul0^#(y, y')
  , mul0^#(C(x, y), y') -> add0^#(mul0(y, y'), y')
  , add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y'))) }
Weak Trs:
  { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
  , mul0(Z(), y) -> Z()
  , add0(C(x, y), y') -> add0(y, C(S(), y'))
  , add0(Z(), y) -> y }
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.

{ mul0^#(C(x, y), y') -> mul0^#(y, y')
, mul0^#(C(x, y), y') -> add0^#(mul0(y, y'), y')
, add0^#(C(x, y), y') -> c_3(add0^#(y, C(S(), y'))) }

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

Weak Trs:
  { mul0(C(x, y), y') -> add0(mul0(y, y'), y')
  , mul0(Z(), y) -> Z()
  , add0(C(x, y), y') -> add0(y, C(S(), y'))
  , add0(Z(), y) -> y }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(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(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^3))