We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ f(x, 0()) -> x
, f(f(x, y), z) -> f(x, f(y, z))
, f(0(), y) -> y
, f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z))
, f(1(), g(x, y)) -> x
, f(2(), g(x, y)) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following dependency tuples:
Strict DPs:
{ f^#(x, 0()) -> c_1()
, f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z))
, f^#(0(), y) -> c_3()
, f^#(i(x), y) -> c_4()
, f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z))
, f^#(1(), g(x, y)) -> c_6()
, f^#(2(), g(x, y)) -> c_7() }
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:
{ f^#(x, 0()) -> c_1()
, f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z))
, f^#(0(), y) -> c_3()
, f^#(i(x), y) -> c_4()
, f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z))
, f^#(1(), g(x, y)) -> c_6()
, f^#(2(), g(x, y)) -> c_7() }
Weak Trs:
{ f(x, 0()) -> x
, f(f(x, y), z) -> f(x, f(y, z))
, f(0(), y) -> y
, f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z))
, f(1(), g(x, y)) -> x
, f(2(), g(x, y)) -> y }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {1,3,4,6,7} by
applications of Pre({1,3,4,6,7}) = {2,5}. Here rules are labeled as
follows:
DPs:
{ 1: f^#(x, 0()) -> c_1()
, 2: f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z))
, 3: f^#(0(), y) -> c_3()
, 4: f^#(i(x), y) -> c_4()
, 5: f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z))
, 6: f^#(1(), g(x, y)) -> c_6()
, 7: f^#(2(), g(x, y)) -> c_7() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z))
, f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z)) }
Weak DPs:
{ f^#(x, 0()) -> c_1()
, f^#(0(), y) -> c_3()
, f^#(i(x), y) -> c_4()
, f^#(1(), g(x, y)) -> c_6()
, f^#(2(), g(x, y)) -> c_7() }
Weak Trs:
{ f(x, 0()) -> x
, f(f(x, y), z) -> f(x, f(y, z))
, f(0(), y) -> y
, f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z))
, f(1(), g(x, y)) -> x
, f(2(), g(x, 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.
{ f^#(x, 0()) -> c_1()
, f^#(0(), y) -> c_3()
, f^#(i(x), y) -> c_4()
, f^#(1(), g(x, y)) -> c_6()
, f^#(2(), g(x, y)) -> c_7() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z))
, f^#(g(x, y), z) -> c_5(f^#(x, z), f^#(y, z)) }
Weak Trs:
{ f(x, 0()) -> x
, f(f(x, y), z) -> f(x, f(y, z))
, f(0(), y) -> y
, f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z))
, f(1(), g(x, y)) -> x
, f(2(), g(x, 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:
{ f^#(f(x, y), z) -> c_2(f^#(x, f(y, z)), f^#(y, z)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ f^#(f(x, y), z) -> c_1(f^#(y, z))
, f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, z)) }
Weak Trs:
{ f(x, 0()) -> x
, f(f(x, y), z) -> f(x, f(y, z))
, f(0(), y) -> y
, f(i(x), y) -> i(x)
, f(g(x, y), z) -> g(f(x, z), f(y, z))
, f(1(), g(x, y)) -> x
, f(2(), g(x, 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:
{ f^#(f(x, y), z) -> c_1(f^#(y, z))
, f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, z)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 1: f^#(f(x, y), z) -> c_1(f^#(y, z))
, 2: f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, z)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[f](x1, x2) = [2] x1 + [2] x2 + [2]
[g](x1, x2) = [1] x1 + [1] x2 + [2]
[f^#](x1, x2) = [4] x1 + [0]
[c_1](x1) = [1] x1 + [1]
[c_2](x1, x2) = [1] x1 + [1] x2 + [0]
The order satisfies the following ordering constraints:
[f^#(f(x, y), z)] = [8] y + [8] x + [8]
> [4] y + [1]
= [c_1(f^#(y, z))]
[f^#(g(x, y), z)] = [4] y + [4] x + [8]
> [4] y + [4] x + [0]
= [c_2(f^#(x, z), f^#(y, 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:
{ f^#(f(x, y), z) -> c_1(f^#(y, z))
, f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, 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.
{ f^#(f(x, y), z) -> c_1(f^#(y, z))
, f^#(g(x, y), z) -> c_2(f^#(x, z), f^#(y, 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))