We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, quot(0(), s(y)) -> 0()
, quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We add the following dependency tuples:
Strict DPs:
{ minus^#(x, 0()) -> c_1()
, minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
, quot^#(0(), s(y)) -> c_3()
, quot^#(s(x), s(y)) ->
c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
minus^#(y, s(s(z))),
minus^#(x, s(0())))
, plus^#(0(), y) -> c_6()
, plus^#(s(x), y) -> c_7(plus^#(x, y))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_8(plus^#(plus(y, s(s(z))), plus(x, s(0()))),
plus^#(y, s(s(z))),
plus^#(x, s(0()))) }
and mark the set of starting terms.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ minus^#(x, 0()) -> c_1()
, minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
, quot^#(0(), s(y)) -> c_3()
, quot^#(s(x), s(y)) ->
c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
minus^#(y, s(s(z))),
minus^#(x, s(0())))
, plus^#(0(), y) -> c_6()
, plus^#(s(x), y) -> c_7(plus^#(x, y))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_8(plus^#(plus(y, s(s(z))), plus(x, s(0()))),
plus^#(y, s(s(z))),
plus^#(x, s(0()))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, quot(0(), s(y)) -> 0()
, quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We estimate the number of application of {1,3,6} by applications of
Pre({1,3,6}) = {2,4,5,7,8}. Here rules are labeled as follows:
DPs:
{ 1: minus^#(x, 0()) -> c_1()
, 2: minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
, 3: quot^#(0(), s(y)) -> c_3()
, 4: quot^#(s(x), s(y)) ->
c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
, 5: plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
minus^#(y, s(s(z))),
minus^#(x, s(0())))
, 6: plus^#(0(), y) -> c_6()
, 7: plus^#(s(x), y) -> c_7(plus^#(x, y))
, 8: plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_8(plus^#(plus(y, s(s(z))), plus(x, s(0()))),
plus^#(y, s(s(z))),
plus^#(x, s(0()))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
, quot^#(s(x), s(y)) ->
c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
minus^#(y, s(s(z))),
minus^#(x, s(0())))
, plus^#(s(x), y) -> c_7(plus^#(x, y))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_8(plus^#(plus(y, s(s(z))), plus(x, s(0()))),
plus^#(y, s(s(z))),
plus^#(x, s(0()))) }
Weak DPs:
{ minus^#(x, 0()) -> c_1()
, quot^#(0(), s(y)) -> c_3()
, plus^#(0(), y) -> c_6() }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, quot(0(), s(y)) -> 0()
, quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ minus^#(x, 0()) -> c_1()
, quot^#(0(), s(y)) -> c_3()
, plus^#(0(), y) -> c_6() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
, quot^#(s(x), s(y)) ->
c_4(quot^#(minus(x, y), s(y)), minus^#(x, y))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
minus^#(y, s(s(z))),
minus^#(x, s(0())))
, plus^#(s(x), y) -> c_7(plus^#(x, y))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_8(plus^#(plus(y, s(s(z))), plus(x, s(0()))),
plus^#(y, s(s(z))),
plus^#(x, s(0()))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, quot(0(), s(y)) -> 0()
, quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
Due to missing edges in the dependency-graph, the right-hand sides
of following rules could be simplified:
{ plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))),
minus^#(y, s(s(z))),
minus^#(x, s(0())))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_8(plus^#(plus(y, s(s(z))), plus(x, s(0()))),
plus^#(y, s(s(z))),
plus^#(x, s(0()))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
, quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(s(x), y) -> c_4(plus^#(x, y))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_5(plus^#(plus(y, s(s(z))), plus(x, s(0())))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, quot(0(), s(y)) -> 0()
, quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
, quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(s(x), y) -> c_4(plus^#(x, y))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_5(plus^#(plus(y, s(s(z))), plus(x, s(0())))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 3: plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_3(plus^#(minus(y, s(s(z))), minus(x, s(0())))) }
Trs: { minus(s(x), s(y)) -> minus(x, y) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_4) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [4] x1 + [5] x2 + [0]
[0] = [0]
[s](x1) = [1] x1 + [1]
[plus](x1, x2) = [1] x1 + [0]
[minus^#](x1, x2) = [0]
[quot^#](x1, x2) = [6]
[plus^#](x1, x2) = [1] x2 + [0]
[c_1](x1) = [1] x1 + [0]
[c_2](x1, x2) = [1] x1 + [4] x2 + [0]
[c_3](x1) = [1]
[c_4](x1) = [1] x1 + [0]
[c_5](x1) = [0]
The order satisfies the following ordering constraints:
[minus(x, 0())] = [4] x + [0]
>= [1] x + [0]
= [x]
[minus(s(x), s(y))] = [4] x + [5] y + [9]
> [4] x + [5] y + [0]
= [minus(x, y)]
[plus(minus(x, s(0())), minus(y, s(s(z))))] = [4] x + [5]
? [4] y + [5] z + [10]
= [plus(minus(y, s(s(z))), minus(x, s(0())))]
[plus(0(), y)] = [0]
? [1] y + [0]
= [y]
[plus(s(x), y)] = [1] x + [1]
>= [1] x + [1]
= [s(plus(x, y))]
[plus(plus(x, s(0())), plus(y, s(s(z))))] = [1] x + [0]
? [1] y + [0]
= [plus(plus(y, s(s(z))), plus(x, s(0())))]
[minus^#(s(x), s(y))] = [0]
>= [0]
= [c_1(minus^#(x, y))]
[quot^#(s(x), s(y))] = [6]
>= [6]
= [c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))]
[plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [4] y + [5] z + [10]
> [1]
= [c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))]
[plus^#(s(x), y)] = [1] y + [0]
>= [1] y + [0]
= [c_4(plus^#(x, y))]
[plus^#(plus(x, s(0())), plus(y, s(s(z))))] = [1] y + [0]
>= [0]
= [c_5(plus^#(plus(y, s(s(z))), plus(x, s(0()))))]
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(n^2)).
Strict DPs:
{ minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
, quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))
, plus^#(s(x), y) -> c_4(plus^#(x, y))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_5(plus^#(plus(y, s(s(z))), plus(x, s(0())))) }
Weak DPs:
{ plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_3(plus^#(minus(y, s(s(z))), minus(x, s(0())))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 4: plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_5(plus^#(plus(y, s(s(z))), plus(x, s(0())))) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_4) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [0]
[0] = [1]
[s](x1) = [1]
[plus](x1, x2) = [2] x2 + [0]
[minus^#](x1, x2) = [0]
[quot^#](x1, x2) = [0]
[plus^#](x1, x2) = [4] x2 + [0]
[c_1](x1) = [4] x1 + [0]
[c_2](x1, x2) = [4] x1 + [1] x2 + [0]
[c_3](x1) = [0]
[c_4](x1) = [1] x1 + [0]
[c_5](x1) = [1]
The order satisfies the following ordering constraints:
[minus(x, 0())] = [0]
? [1] x + [0]
= [x]
[minus(s(x), s(y))] = [0]
>= [0]
= [minus(x, y)]
[plus(minus(x, s(0())), minus(y, s(s(z))))] = [0]
>= [0]
= [plus(minus(y, s(s(z))), minus(x, s(0())))]
[plus(0(), y)] = [2] y + [0]
>= [1] y + [0]
= [y]
[plus(s(x), y)] = [2] y + [0]
? [1]
= [s(plus(x, y))]
[plus(plus(x, s(0())), plus(y, s(s(z))))] = [4]
>= [4]
= [plus(plus(y, s(s(z))), plus(x, s(0())))]
[minus^#(s(x), s(y))] = [0]
>= [0]
= [c_1(minus^#(x, y))]
[quot^#(s(x), s(y))] = [0]
>= [0]
= [c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))]
[plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [0]
>= [0]
= [c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))]
[plus^#(s(x), y)] = [4] y + [0]
>= [4] y + [0]
= [c_4(plus^#(x, y))]
[plus^#(plus(x, s(0())), plus(y, s(s(z))))] = [8]
> [1]
= [c_5(plus^#(plus(y, s(s(z))), plus(x, s(0()))))]
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(n^2)).
Strict DPs:
{ minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
, quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))
, plus^#(s(x), y) -> c_4(plus^#(x, y)) }
Weak DPs:
{ plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_5(plus^#(plus(y, s(s(z))), plus(x, s(0())))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'matrix interpretation of dimension 1' to
orient following rules strictly.
DPs:
{ 2: quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))
, 3: plus^#(s(x), y) -> c_4(plus^#(x, y))
, 4: plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, 5: plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_5(plus^#(plus(y, s(s(z))), plus(x, s(0())))) }
Trs:
{ minus(s(x), s(y)) -> minus(x, y)
, plus(s(x), y) -> s(plus(x, y)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_4) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [1] x1 + [0]
[0] = [0]
[s](x1) = [1] x1 + [1]
[plus](x1, x2) = [3] x1 + [0]
[minus^#](x1, x2) = [0]
[quot^#](x1, x2) = [7] x1 + [7]
[plus^#](x1, x2) = [1] x1 + [1]
[c_1](x1) = [1] x1 + [0]
[c_2](x1, x2) = [1] x1 + [1] x2 + [1]
[c_3](x1) = [0]
[c_4](x1) = [1] x1 + [0]
[c_5](x1) = [0]
The order satisfies the following ordering constraints:
[minus(x, 0())] = [1] x + [0]
>= [1] x + [0]
= [x]
[minus(s(x), s(y))] = [1] x + [1]
> [1] x + [0]
= [minus(x, y)]
[plus(minus(x, s(0())), minus(y, s(s(z))))] = [3] x + [0]
? [3] y + [0]
= [plus(minus(y, s(s(z))), minus(x, s(0())))]
[plus(0(), y)] = [0]
? [1] y + [0]
= [y]
[plus(s(x), y)] = [3] x + [3]
> [3] x + [1]
= [s(plus(x, y))]
[plus(plus(x, s(0())), plus(y, s(s(z))))] = [9] x + [0]
? [9] y + [0]
= [plus(plus(y, s(s(z))), plus(x, s(0())))]
[minus^#(s(x), s(y))] = [0]
>= [0]
= [c_1(minus^#(x, y))]
[quot^#(s(x), s(y))] = [7] x + [14]
> [7] x + [8]
= [c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))]
[plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [1] x + [1]
> [0]
= [c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))]
[plus^#(s(x), y)] = [1] x + [2]
> [1] x + [1]
= [c_4(plus^#(x, y))]
[plus^#(plus(x, s(0())), plus(y, s(s(z))))] = [3] x + [1]
> [0]
= [c_5(plus^#(plus(y, s(s(z))), plus(x, s(0()))))]
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(n^2)).
Strict DPs: { minus^#(s(x), s(y)) -> c_1(minus^#(x, y)) }
Weak DPs:
{ quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(s(x), y) -> c_4(plus^#(x, y))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_5(plus^#(plus(y, s(s(z))), plus(x, s(0())))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
The following weak DPs constitute a sub-graph of the DG that is
closed under successors. The DPs are removed.
{ plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_3(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(s(x), y) -> c_4(plus^#(x, y))
, plus^#(plus(x, s(0())), plus(y, s(s(z)))) ->
c_5(plus^#(plus(y, s(s(z))), plus(x, s(0())))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs: { minus^#(s(x), s(y)) -> c_1(minus^#(x, y)) }
Weak DPs:
{ quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, plus(minus(x, s(0())), minus(y, s(s(z)))) ->
plus(minus(y, s(s(z))), minus(x, s(0())))
, plus(0(), y) -> y
, plus(s(x), y) -> s(plus(x, y))
, plus(plus(x, s(0())), plus(y, s(s(z)))) ->
plus(plus(y, s(s(z))), plus(x, s(0()))) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs: { minus^#(s(x), s(y)) -> c_1(minus^#(x, y)) }
Weak DPs:
{ quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We use the processor 'Small Polynomial Path Order (PS,2-bounded)'
to orient following rules strictly.
DPs:
{ 1: minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
, 2: quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) }
Trs: { minus(s(x), s(y)) -> minus(x, y) }
Sub-proof:
----------
The input was oriented with the instance of 'Small Polynomial Path
Order (PS,2-bounded)' as induced by the safe mapping
safe(minus) = {1}, safe(0) = {}, safe(s) = {1}, safe(plus) = {},
safe(minus^#) = {}, safe(quot^#) = {}, safe(plus^#) = {},
safe(c_1) = {}, safe(c_2) = {}, safe(c_3) = {}, safe(c_4) = {},
safe(c_5) = {}
and precedence
quot^# > minus^# .
Following symbols are considered recursive:
{minus^#, quot^#}
The recursion depth is 2.
Further, following argument filtering is employed:
pi(minus) = 1, pi(0) = [], pi(s) = [1], pi(plus) = [],
pi(minus^#) = [1, 2], pi(quot^#) = [1, 2], pi(plus^#) = [],
pi(c_1) = [1], pi(c_2) = [1, 2], pi(c_3) = [], pi(c_4) = [],
pi(c_5) = []
Usable defined function symbols are a subset of:
{minus, minus^#, quot^#, plus^#}
For your convenience, here are the satisfied ordering constraints:
pi(minus^#(s(x), s(y))) = minus^#(s(; x), s(; y);)
> c_1(minus^#(x, y;);)
= pi(c_1(minus^#(x, y)))
pi(quot^#(s(x), s(y))) = quot^#(s(; x), s(; y);)
> c_2(quot^#(x, s(; y);), minus^#(x, y;);)
= pi(c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)))
pi(minus(x, 0())) = x
>= x
= pi(x)
pi(minus(s(x), s(y))) = s(; x)
> x
= pi(minus(x, 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:
{ minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
, quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, 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.
{ minus^#(s(x), s(y)) -> c_1(minus^#(x, y))
, quot^#(s(x), s(y)) ->
c_2(quot^#(minus(x, y), s(y)), minus^#(x, y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, 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^2))