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:
  { f(a(), a()) -> f(a(), b())
  , f(a(), b()) -> f(s(a()), c())
  , f(s(X), c()) -> f(X, c())
  , f(c(), c()) -> f(a(), a()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs: { f(c(), c()) -> f(a(), a()) }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
    [f](x1, x2) = [2 2] x1 + [2 0] x2 + [0]
                  [2 2]      [0 0]      [0]
                                           
            [a] = [0]                      
                  [0]                      
                                           
            [b] = [0]                      
                  [0]                      
                                           
        [s](x1) = [1 2] x1 + [0]           
                  [0 0]      [0]           
                                           
            [c] = [0]                      
                  [2]                      
  
  This order satisfies the following ordering constraints:
  
     [f(a(), a())] =  [0]             
                      [0]             
                   >= [0]             
                      [0]             
                   =  [f(a(), b())]   
                                      
     [f(a(), b())] =  [0]             
                      [0]             
                   >= [0]             
                      [0]             
                   =  [f(s(a()), c())]
                                      
    [f(s(X), c())] =  [2 4] X + [0]   
                      [2 4]     [0]   
                   >= [2 2] X + [0]   
                      [2 2]     [0]   
                   =  [f(X, c())]     
                                      
     [f(c(), c())] =  [4]             
                      [4]             
                   >  [0]             
                      [0]             
                   =  [f(a(), a())]   
                                      

We return to the main proof.

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

Strict Trs:
  { f(a(), a()) -> f(a(), b())
  , f(a(), b()) -> f(s(a()), c())
  , f(s(X), c()) -> f(X, c()) }
Weak Trs: { f(c(), c()) -> f(a(), a()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs: { f(a(), a()) -> f(a(), b()) }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
    [f](x1, x2) = [1 1] x1 + [2 0] x2 + [0]
                  [0 0]      [0 0]      [0]
                                           
            [a] = [1]                      
                  [0]                      
                                           
            [b] = [0]                      
                  [0]                      
                                           
        [s](x1) = [1 2] x1 + [0]           
                  [0 0]      [0]           
                                           
            [c] = [0]                      
                  [3]                      
  
  This order satisfies the following ordering constraints:
  
     [f(a(), a())] =  [3]             
                      [0]             
                   >  [1]             
                      [0]             
                   =  [f(a(), b())]   
                                      
     [f(a(), b())] =  [1]             
                      [0]             
                   >= [1]             
                      [0]             
                   =  [f(s(a()), c())]
                                      
    [f(s(X), c())] =  [1 2] X + [0]   
                      [0 0]     [0]   
                   >= [1 1] X + [0]   
                      [0 0]     [0]   
                   =  [f(X, c())]     
                                      
     [f(c(), c())] =  [3]             
                      [0]             
                   >= [3]             
                      [0]             
                   =  [f(a(), a())]   
                                      

We return to the main proof.

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

Strict Trs:
  { f(a(), b()) -> f(s(a()), c())
  , f(s(X), c()) -> f(X, c()) }
Weak Trs:
  { f(a(), a()) -> f(a(), b())
  , f(c(), c()) -> f(a(), a()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs: { f(a(), b()) -> f(s(a()), c()) }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
    [f](x1, x2) = [1 0] x1 + [1 2] x2 + [0]
                  [0 0]      [0 0]      [0]
                                           
            [a] = [0]                      
                  [2]                      
                                           
            [b] = [0]                      
                  [2]                      
                                           
        [s](x1) = [1 0] x1 + [0]           
                  [0 0]      [0]           
                                           
            [c] = [1]                      
                  [1]                      
  
  This order satisfies the following ordering constraints:
  
     [f(a(), a())] =  [4]             
                      [0]             
                   >= [4]             
                      [0]             
                   =  [f(a(), b())]   
                                      
     [f(a(), b())] =  [4]             
                      [0]             
                   >  [3]             
                      [0]             
                   =  [f(s(a()), c())]
                                      
    [f(s(X), c())] =  [1 0] X + [3]   
                      [0 0]     [0]   
                   >= [1 0] X + [3]   
                      [0 0]     [0]   
                   =  [f(X, c())]     
                                      
     [f(c(), c())] =  [4]             
                      [0]             
                   >= [4]             
                      [0]             
                   =  [f(a(), a())]   
                                      

We return to the main proof.

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

Strict Trs: { f(s(X), c()) -> f(X, c()) }
Weak Trs:
  { f(a(), a()) -> f(a(), b())
  , f(a(), b()) -> f(s(a()), c())
  , f(c(), c()) -> f(a(), a()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(n^1))

We use the processor 'matrix interpretation of dimension 2' to
orient following rules strictly.

Trs: { f(s(X), c()) -> f(X, c()) }

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:
----------
  TcT has computed the following constructor-based matrix
  interpretation satisfying not(EDA) and not(IDA(1)).
  
    [f](x1, x2) = [1 2] x1 + [2 0] x2 + [0]
                  [0 0]      [0 0]      [0]
                                           
            [a] = [2]                      
                  [0]                      
                                           
            [b] = [2]                      
                  [0]                      
                                           
        [s](x1) = [1 2] x1 + [0]           
                  [0 0]      [2]           
                                           
            [c] = [0]                      
                  [3]                      
  
  This order satisfies the following ordering constraints:
  
     [f(a(), a())] =  [6]             
                      [0]             
                   >= [6]             
                      [0]             
                   =  [f(a(), b())]   
                                      
     [f(a(), b())] =  [6]             
                      [0]             
                   >= [6]             
                      [0]             
                   =  [f(s(a()), c())]
                                      
    [f(s(X), c())] =  [1 2] X + [4]   
                      [0 0]     [0]   
                   >  [1 2] X + [0]   
                      [0 0]     [0]   
                   =  [f(X, c())]     
                                      
     [f(c(), c())] =  [6]             
                      [0]             
                   >= [6]             
                      [0]             
                   =  [f(a(), a())]   
                                      

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:
  { f(a(), a()) -> f(a(), b())
  , f(a(), b()) -> f(s(a()), c())
  , f(s(X), c()) -> f(X, c())
  , f(c(), c()) -> f(a(), a()) }
Obligation:
  innermost runtime complexity
Answer:
  YES(O(1),O(1))

Empty rules are trivially bounded

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