We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
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)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following weak dependency pairs:
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)))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(0(), y) -> c_6()
, plus^#(s(x), y) -> c_7(plus^#(x, y)) }
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:
{ 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)))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(0(), y) -> c_6()
, plus^#(s(x), y) -> c_7(plus^#(x, y)) }
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)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We replace rewrite rules by usable rules:
Strict 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^1)).
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)))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(0(), y) -> c_6()
, plus^#(s(x), y) -> c_7(plus^#(x, y)) }
Strict Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
The weightgap principle applies (using the following constant
growth matrix-interpretation)
The following argument positions are usable:
Uargs(c_2) = {1}, Uargs(quot^#) = {1}, Uargs(c_4) = {1},
Uargs(c_7) = {1}
TcT has computed the following constructor-restricted matrix
interpretation.
[minus](x1, x2) = [1 1] x1 + [2]
[2 2] [0]
[0] = [0]
[0]
[s](x1) = [1 1] x1 + [0]
[0 0] [2]
[minus^#](x1, x2) = [0]
[0]
[c_1] = [0]
[0]
[c_2](x1) = [1 0] x1 + [0]
[0 1] [0]
[quot^#](x1, x2) = [1 0] x1 + [0]
[0 0] [0]
[c_3] = [0]
[0]
[c_4](x1) = [1 0] x1 + [0]
[0 1] [0]
[plus^#](x1, x2) = [0]
[0]
[c_5](x1) = [0]
[0]
[c_6] = [0]
[0]
[c_7](x1) = [1 0] x1 + [0]
[0 1] [0]
The order satisfies the following ordering constraints:
[minus(x, 0())] = [1 1] x + [2]
[2 2] [0]
> [1 0] x + [0]
[0 1] [0]
= [x]
[minus(s(x), s(y))] = [1 1] x + [4]
[2 2] [4]
> [1 1] x + [2]
[2 2] [0]
= [minus(x, y)]
[minus^#(x, 0())] = [0]
[0]
>= [0]
[0]
= [c_1()]
[minus^#(s(x), s(y))] = [0]
[0]
>= [0]
[0]
= [c_2(minus^#(x, y))]
[quot^#(0(), s(y))] = [0]
[0]
>= [0]
[0]
= [c_3()]
[quot^#(s(x), s(y))] = [1 1] x + [0]
[0 0] [0]
? [1 1] x + [2]
[0 0] [0]
= [c_4(quot^#(minus(x, y), s(y)))]
[plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [0]
[0]
>= [0]
[0]
= [c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))]
[plus^#(0(), y)] = [0]
[0]
>= [0]
[0]
= [c_6()]
[plus^#(s(x), y)] = [0]
[0]
>= [0]
[0]
= [c_7(plus^#(x, y))]
Further, it can be verified that all rules not oriented are covered by the weightgap condition.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
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)))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(0(), y) -> c_6()
, plus^#(s(x), y) -> c_7(plus^#(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^1))
We estimate the number of application of {1,3,6} by applications of
Pre({1,3,6}) = {2,4,5,7}. 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)))
, 5: plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(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)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
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)))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(s(x), y) -> c_7(plus^#(x, y)) }
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) }
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.
{ 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^1)).
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)))
, plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(s(x), y) -> c_7(plus^#(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^1))
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_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_2) = {1}, Uargs(c_4) = {1}, Uargs(c_7) = {1}
TcT has computed the following constructor-restricted matrix
interpretation. Note that the diagonal of the component-wise maxima
of interpretation-entries (of constructors) contains no more than 0
non-zero entries.
[minus](x1, x2) = [0]
[0] = [0]
[s](x1) = [0]
[minus^#](x1, x2) = [0]
[c_2](x1) = [4] x1 + [0]
[quot^#](x1, x2) = [0]
[c_4](x1) = [4] x1 + [0]
[plus^#](x1, x2) = [1] x2 + [1]
[c_5](x1) = [0]
[c_7](x1) = [1] x1 + [0]
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)]
[minus^#(s(x), s(y))] = [0]
>= [0]
= [c_2(minus^#(x, y))]
[quot^#(s(x), s(y))] = [0]
>= [0]
= [c_4(quot^#(minus(x, y), s(y)))]
[plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [1]
> [0]
= [c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))]
[plus^#(s(x), y)] = [1] y + [1]
>= [1] y + [1]
= [c_7(plus^#(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(n^1)).
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)))
, plus^#(s(x), y) -> c_7(plus^#(x, y)) }
Weak DPs:
{ plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y) }
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: minus^#(s(x), s(y)) -> c_2(minus^#(x, y))
, 4: plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_2) = {1}, Uargs(c_4) = {1}, Uargs(c_7) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [0]
[0] = [0]
[s](x1) = [1] x1 + [4]
[minus^#](x1, x2) = [2] x1 + [0]
[c_2](x1) = [1] x1 + [1]
[quot^#](x1, x2) = [0]
[c_4](x1) = [1] x1 + [0]
[plus^#](x1, x2) = [1] x2 + [4]
[c_5](x1) = [0]
[c_7](x1) = [1] x1 + [0]
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)]
[minus^#(s(x), s(y))] = [2] x + [8]
> [2] x + [1]
= [c_2(minus^#(x, y))]
[quot^#(s(x), s(y))] = [0]
>= [0]
= [c_4(quot^#(minus(x, y), s(y)))]
[plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [4]
> [0]
= [c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))]
[plus^#(s(x), y)] = [1] y + [4]
>= [1] y + [4]
= [c_7(plus^#(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(n^1)).
Strict DPs:
{ quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y)))
, plus^#(s(x), y) -> c_7(plus^#(x, y)) }
Weak DPs:
{ minus^#(s(x), s(y)) -> c_2(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())))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, 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.
{ minus^#(s(x), s(y)) -> c_2(minus^#(x, y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y)))
, plus^#(s(x), y) -> c_7(plus^#(x, y)) }
Weak DPs:
{ plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0())))) }
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y) }
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:
{ 2: plus^#(s(x), y) -> c_7(plus^#(x, y)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_4) = {1}, Uargs(c_7) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[minus](x1, x2) = [0]
[0] = [0]
[s](x1) = [1] x1 + [5]
[minus^#](x1, x2) = [0]
[c_2](x1) = [0]
[quot^#](x1, x2) = [2]
[c_4](x1) = [1] x1 + [0]
[plus^#](x1, x2) = [1] x1 + [1] x2 + [0]
[c_5](x1) = [0]
[c_7](x1) = [1] x1 + [0]
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)]
[quot^#(s(x), s(y))] = [2]
>= [2]
= [c_4(quot^#(minus(x, y), s(y)))]
[plus^#(minus(x, s(0())), minus(y, s(s(z))))] = [0]
>= [0]
= [c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))]
[plus^#(s(x), y)] = [1] x + [1] y + [5]
> [1] x + [1] y + [0]
= [c_7(plus^#(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(n^1)).
Strict DPs:
{ quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(y))) }
Weak DPs:
{ plus^#(minus(x, s(0())), minus(y, s(s(z)))) ->
c_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(s(x), y) -> c_7(plus^#(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^1))
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_5(plus^#(minus(y, s(s(z))), minus(x, s(0()))))
, plus^#(s(x), y) -> c_7(plus^#(x, y)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(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^1))
We use the processor 'Small Polynomial Path Order (PS,1-bounded)'
to orient following rules strictly.
DPs:
{ 1: quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(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,1-bounded)' as induced by the safe mapping
safe(minus) = {1, 2}, safe(0) = {}, safe(s) = {1},
safe(minus^#) = {}, safe(c_2) = {}, safe(quot^#) = {2},
safe(c_4) = {}, safe(plus^#) = {}, safe(c_5) = {}, safe(c_7) = {}
and precedence
empty .
Following symbols are considered recursive:
{quot^#}
The recursion depth is 1.
Further, following argument filtering is employed:
pi(minus) = 1, pi(0) = [], pi(s) = [1], pi(minus^#) = [],
pi(c_2) = [], pi(quot^#) = [1, 2], pi(c_4) = [1], pi(plus^#) = [],
pi(c_5) = [], pi(c_7) = []
Usable defined function symbols are a subset of:
{minus, minus^#, quot^#, plus^#}
For your convenience, here are the satisfied ordering constraints:
pi(quot^#(s(x), s(y))) = quot^#(s(; x); s(; y))
> c_4(quot^#(x; s(; y));)
= pi(c_4(quot^#(minus(x, y), s(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: { quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(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.
{ quot^#(s(x), s(y)) -> c_4(quot^#(minus(x, y), s(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^1))