We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We add the following dependency tuples:
Strict DPs:
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, active^#(f(a(), X, X)) -> c_2(f^#(X, b(), b()))
, active^#(b()) -> c_3()
, f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3))
, proper^#(a()) -> c_7()
, proper^#(b()) -> c_8()
, top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
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:
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, active^#(f(a(), X, X)) -> c_2(f^#(X, b(), b()))
, active^#(b()) -> c_3()
, f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3))
, proper^#(a()) -> c_7()
, proper^#(b()) -> c_8()
, top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We estimate the number of application of {2,3,7,8} by applications
of Pre({2,3,7,8}) = {1,6,9,10}. Here rules are labeled as follows:
DPs:
{ 1: active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, 2: active^#(f(a(), X, X)) -> c_2(f^#(X, b(), b()))
, 3: active^#(b()) -> c_3()
, 4: f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, 5: f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, 6: proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3))
, 7: proper^#(a()) -> c_7()
, 8: proper^#(b()) -> c_8()
, 9: top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, 10: top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3))
, top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
Weak DPs:
{ active^#(f(a(), X, X)) -> c_2(f^#(X, b(), b()))
, active^#(b()) -> c_3()
, proper^#(a()) -> c_7()
, proper^#(b()) -> c_8() }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X)) }
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.
{ active^#(f(a(), X, X)) -> c_2(f^#(X, b(), b()))
, active^#(b()) -> c_3()
, proper^#(a()) -> c_7()
, proper^#(b()) -> c_8() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3))
, top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b())
, top(mark(X)) -> top(proper(X))
, top(ok(X)) -> top(active(X)) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We replace rewrite rules by usable rules:
Weak Usable Rules:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^2)).
Strict DPs:
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3))
, top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^2))
We decompose the input problem according to the dependency graph
into the upper component
{ top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
and lower component
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3)) }
Further, following extension rules are added to the lower
component.
{ top^#(mark(X)) -> proper^#(X)
, top^#(mark(X)) -> top^#(proper(X))
, top^#(ok(X)) -> active^#(X)
, top^#(ok(X)) -> top^#(active(X)) }
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:
{ top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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.
DPs:
{ 1: top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X)) }
Trs:
{ active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a()) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_9) = {1}, Uargs(c_10) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[active](x1) = [0 2] x1 + [0]
[0 1] [0]
[f](x1, x2, x3) = [1 3] x1 + [1 0] x2 + [1 0] x3 + [0]
[1 2] [0 1] [4 1] [0]
[a] = [6]
[0]
[mark](x1) = [0 0] x1 + [3]
[0 1] [2]
[b] = [0]
[2]
[proper](x1) = [1 0] x1 + [0]
[0 1] [0]
[ok](x1) = [1 0] x1 + [0]
[0 1] [0]
[active^#](x1) = [0 1] x1 + [1]
[0 0] [1]
[proper^#](x1) = [2]
[1]
[top^#](x1) = [0 7] x1 + [0]
[1 1] [0]
[c_9](x1, x2) = [1 0] x1 + [1]
[0 0] [0]
[c_10](x1, x2) = [1 0] x1 + [0]
[0 0] [0]
The order satisfies the following ordering constraints:
[active(f(X1, X2, X3))] = [2 4] X1 + [0 2] X2 + [8 2] X3 + [0]
[1 2] [0 1] [4 1] [0]
>= [1 3] X1 + [0 2] X2 + [1 0] X3 + [0]
[1 2] [0 1] [4 1] [0]
= [f(X1, active(X2), X3)]
[active(f(a(), X, X))] = [8 4] X + [12]
[4 2] [6]
> [0 0] X + [3]
[1 2] [6]
= [mark(f(X, b(), b()))]
[active(b())] = [4]
[2]
> [3]
[2]
= [mark(a())]
[f(X1, mark(X2), X3)] = [1 3] X1 + [0 0] X2 + [1 0] X3 + [3]
[1 2] [0 1] [4 1] [2]
>= [0 0] X1 + [0 0] X2 + [0 0] X3 + [3]
[1 2] [0 1] [4 1] [2]
= [mark(f(X1, X2, X3))]
[f(ok(X1), ok(X2), ok(X3))] = [1 3] X1 + [1 0] X2 + [1 0] X3 + [0]
[1 2] [0 1] [4 1] [0]
>= [1 3] X1 + [1 0] X2 + [1 0] X3 + [0]
[1 2] [0 1] [4 1] [0]
= [ok(f(X1, X2, X3))]
[proper(f(X1, X2, X3))] = [1 3] X1 + [1 0] X2 + [1 0] X3 + [0]
[1 2] [0 1] [4 1] [0]
>= [1 3] X1 + [1 0] X2 + [1 0] X3 + [0]
[1 2] [0 1] [4 1] [0]
= [f(proper(X1), proper(X2), proper(X3))]
[proper(a())] = [6]
[0]
>= [6]
[0]
= [ok(a())]
[proper(b())] = [0]
[2]
>= [0]
[2]
= [ok(b())]
[top^#(mark(X))] = [0 7] X + [14]
[0 1] [5]
> [0 7] X + [1]
[0 0] [0]
= [c_9(top^#(proper(X)), proper^#(X))]
[top^#(ok(X))] = [0 7] X + [0]
[1 1] [0]
>= [0 7] X + [0]
[0 0] [0]
= [c_10(top^#(active(X)), active^#(X))]
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: { top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
Weak DPs: { top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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.
DPs:
{ 1: top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_9) = {1}, Uargs(c_10) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[active](x1) = [0 0] x1 + [0]
[1 1] [1]
[f](x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[2 0] [0 1] [1 0] [1]
[a] = [2]
[0]
[mark](x1) = [0 0] x1 + [0]
[1 1] [2]
[b] = [0]
[3]
[proper](x1) = [1 0] x1 + [0]
[0 1] [2]
[ok](x1) = [1 0] x1 + [0]
[0 1] [2]
[active^#](x1) = [0]
[1]
[proper^#](x1) = [0]
[0]
[top^#](x1) = [6 6] x1 + [0]
[0 0] [5]
[c_9](x1, x2) = [1 0] x1 + [0]
[0 0] [0]
[c_10](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
The order satisfies the following ordering constraints:
[active(f(X1, X2, X3))] = [0 0] X1 + [0 0] X2 + [0 0] X3 + [0]
[2 0] [1 1] [1 0] [2]
>= [0 0] X1 + [0 0] X2 + [0 0] X3 + [0]
[2 0] [1 1] [1 0] [2]
= [f(X1, active(X2), X3)]
[active(f(a(), X, X))] = [0 0] X + [0]
[2 1] [6]
>= [0 0] X + [0]
[2 0] [6]
= [mark(f(X, b(), b()))]
[active(b())] = [0]
[4]
>= [0]
[4]
= [mark(a())]
[f(X1, mark(X2), X3)] = [0 0] X1 + [0 0] X2 + [0 0] X3 + [0]
[2 0] [1 1] [1 0] [3]
>= [0 0] X1 + [0 0] X2 + [0 0] X3 + [0]
[2 0] [1 1] [1 0] [3]
= [mark(f(X1, X2, X3))]
[f(ok(X1), ok(X2), ok(X3))] = [0 0] X1 + [1 0] X2 + [0 0] X3 + [0]
[2 0] [0 1] [1 0] [3]
>= [0 0] X1 + [1 0] X2 + [0 0] X3 + [0]
[2 0] [0 1] [1 0] [3]
= [ok(f(X1, X2, X3))]
[proper(f(X1, X2, X3))] = [0 0] X1 + [1 0] X2 + [0 0] X3 + [0]
[2 0] [0 1] [1 0] [3]
>= [0 0] X1 + [1 0] X2 + [0 0] X3 + [0]
[2 0] [0 1] [1 0] [3]
= [f(proper(X1), proper(X2), proper(X3))]
[proper(a())] = [2]
[2]
>= [2]
[2]
= [ok(a())]
[proper(b())] = [0]
[5]
>= [0]
[5]
= [ok(b())]
[top^#(mark(X))] = [6 6] X + [12]
[0 0] [5]
>= [6 6] X + [12]
[0 0] [0]
= [c_9(top^#(proper(X)), proper^#(X))]
[top^#(ok(X))] = [6 6] X + [12]
[0 0] [5]
> [6 6] X + [6]
[0 0] [0]
= [c_10(top^#(active(X)), active^#(X))]
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:
{ top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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.
{ top^#(mark(X)) -> c_9(top^#(proper(X)), proper^#(X))
, top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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^1)).
Strict DPs:
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3)) }
Weak DPs:
{ top^#(mark(X)) -> proper^#(X)
, top^#(mark(X)) -> top^#(proper(X))
, top^#(ok(X)) -> active^#(X)
, top^#(ok(X)) -> top^#(active(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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.
DPs:
{ 2: f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, 5: top^#(mark(X)) -> proper^#(X)
, 7: top^#(ok(X)) -> active^#(X) }
Trs:
{ active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1, 2}, Uargs(c_4) = {1}, Uargs(c_5) = {1},
Uargs(c_6) = {1, 2, 3, 4}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[active](x1) = [0 1] x1 + [0]
[0 1] [0]
[f](x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 5] [0 4] [0 1] [0]
[a] = [0]
[3]
[mark](x1) = [1 0] x1 + [2]
[0 1] [0]
[b] = [0]
[3]
[proper](x1) = [0 3] x1 + [0]
[0 1] [0]
[ok](x1) = [0 0] x1 + [0]
[1 1] [0]
[active^#](x1) = [0 3] x1 + [0]
[0 1] [0]
[c_1](x1, x2) = [1 0] x1 + [1 1] x2 + [0]
[0 0] [0 0] [0]
[f^#](x1, x2, x3) = [1 1] x2 + [0 0] x3 + [0]
[0 0] [4 0] [1]
[c_4](x1) = [1 0] x1 + [1]
[0 0] [0]
[c_5](x1) = [1 0] x1 + [0]
[0 0] [0]
[proper^#](x1) = [0 2] x1 + [0]
[0 0] [0]
[c_6](x1, x2, x3, x4) = [1 0] x1 + [4 0] x2 + [1 0] x3 + [1
1] x4 + [0]
[0 0] [0 0] [0 0] [0
0] [0]
[top^#](x1) = [0 4] x1 + [7]
[0 1] [0]
The order satisfies the following ordering constraints:
[active(f(X1, X2, X3))] = [0 5] X1 + [0 4] X2 + [0 1] X3 + [0]
[0 5] [0 4] [0 1] [0]
>= [0 0] X1 + [0 1] X2 + [0 0] X3 + [0]
[0 5] [0 4] [0 1] [0]
= [f(X1, active(X2), X3)]
[active(f(a(), X, X))] = [0 5] X + [15]
[0 5] [15]
> [0 0] X + [2]
[0 5] [15]
= [mark(f(X, b(), b()))]
[active(b())] = [3]
[3]
> [2]
[3]
= [mark(a())]
[f(X1, mark(X2), X3)] = [0 0] X1 + [1 0] X2 + [0 0] X3 + [2]
[0 5] [0 4] [0 1] [0]
>= [0 0] X1 + [1 0] X2 + [0 0] X3 + [2]
[0 5] [0 4] [0 1] [0]
= [mark(f(X1, X2, X3))]
[f(ok(X1), ok(X2), ok(X3))] = [0 0] X1 + [0 0] X2 + [0 0] X3 + [0]
[5 5] [4 4] [1 1] [0]
>= [0 0] X1 + [0 0] X2 + [0 0] X3 + [0]
[0 5] [1 4] [0 1] [0]
= [ok(f(X1, X2, X3))]
[proper(f(X1, X2, X3))] = [0 15] X1 + [0 12] X2 + [0 3] X3 + [0]
[0 5] [0 4] [0 1] [0]
>= [0 0] X1 + [0 3] X2 + [0 0] X3 + [0]
[0 5] [0 4] [0 1] [0]
= [f(proper(X1), proper(X2), proper(X3))]
[proper(a())] = [9]
[3]
> [0]
[3]
= [ok(a())]
[proper(b())] = [9]
[3]
> [0]
[3]
= [ok(b())]
[active^#(f(X1, X2, X3))] = [0 15] X1 + [0 12] X2 + [0 3] X3 + [0]
[0 5] [0 4] [0 1] [0]
>= [0 6] X2 + [0]
[0 0] [0]
= [c_1(f^#(X1, active(X2), X3), active^#(X2))]
[f^#(X1, mark(X2), X3)] = [1 1] X2 + [0 0] X3 + [2]
[0 0] [4 0] [1]
> [1 1] X2 + [1]
[0 0] [0]
= [c_4(f^#(X1, X2, X3))]
[f^#(ok(X1), ok(X2), ok(X3))] = [1 1] X2 + [0]
[0 0] [1]
>= [1 1] X2 + [0]
[0 0] [0]
= [c_5(f^#(X1, X2, X3))]
[proper^#(f(X1, X2, X3))] = [0 10] X1 + [0 8] X2 + [0 2] X3 + [0]
[0 0] [0 0] [0 0] [0]
>= [0 8] X1 + [0 6] X2 + [0 2] X3 + [0]
[0 0] [0 0] [0 0] [0]
= [c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3))]
[top^#(mark(X))] = [0 4] X + [7]
[0 1] [0]
> [0 2] X + [0]
[0 0] [0]
= [proper^#(X)]
[top^#(mark(X))] = [0 4] X + [7]
[0 1] [0]
>= [0 4] X + [7]
[0 1] [0]
= [top^#(proper(X))]
[top^#(ok(X))] = [4 4] X + [7]
[1 1] [0]
> [0 3] X + [0]
[0 1] [0]
= [active^#(X)]
[top^#(ok(X))] = [4 4] X + [7]
[1 1] [0]
>= [0 4] X + [7]
[0 1] [0]
= [top^#(active(X))]
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:
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3)) }
Weak DPs:
{ f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, top^#(mark(X)) -> proper^#(X)
, top^#(mark(X)) -> top^#(proper(X))
, top^#(ok(X)) -> active^#(X)
, top^#(ok(X)) -> top^#(active(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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.
DPs:
{ 1: active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, 2: f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1, 2}, Uargs(c_4) = {1}, Uargs(c_5) = {1},
Uargs(c_6) = {1, 2, 3, 4}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA) and not(IDA(1)).
[active](x1) = [1 0] x1 + [0]
[0 1] [1]
[f](x1, x2, x3) = [5 0] x1 + [1 0] x2 + [4 0] x3 + [1]
[0 1] [0 1] [0 0] [0]
[a] = [1]
[0]
[mark](x1) = [1 0] x1 + [0]
[0 1] [0]
[b] = [1]
[0]
[proper](x1) = [1 0] x1 + [0]
[3 0] [3]
[ok](x1) = [1 2] x1 + [0]
[0 0] [4]
[active^#](x1) = [1 6] x1 + [0]
[0 0] [0]
[c_1](x1, x2) = [2 1] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[f^#](x1, x2, x3) = [1 1] x1 + [0]
[1 0] [0]
[c_4](x1) = [1 0] x1 + [0]
[0 0] [0]
[c_5](x1) = [1 0] x1 + [3]
[0 0] [0]
[proper^#](x1) = [3 0] x1 + [0]
[1 0] [0]
[c_6](x1, x2, x3, x4) = [1 4] x1 + [2 0] x2 + [1 0] x3 + [1
1] x4 + [0]
[0 0] [0 0] [0 0] [0
0] [0]
[top^#](x1) = [3 0] x1 + [0]
[4 0] [4]
The order satisfies the following ordering constraints:
[active(f(X1, X2, X3))] = [5 0] X1 + [1 0] X2 + [4 0] X3 + [1]
[0 1] [0 1] [0 0] [1]
>= [5 0] X1 + [1 0] X2 + [4 0] X3 + [1]
[0 1] [0 1] [0 0] [1]
= [f(X1, active(X2), X3)]
[active(f(a(), X, X))] = [5 0] X + [6]
[0 1] [1]
>= [5 0] X + [6]
[0 1] [0]
= [mark(f(X, b(), b()))]
[active(b())] = [1]
[1]
>= [1]
[0]
= [mark(a())]
[f(X1, mark(X2), X3)] = [5 0] X1 + [1 0] X2 + [4 0] X3 + [1]
[0 1] [0 1] [0 0] [0]
>= [5 0] X1 + [1 0] X2 + [4 0] X3 + [1]
[0 1] [0 1] [0 0] [0]
= [mark(f(X1, X2, X3))]
[f(ok(X1), ok(X2), ok(X3))] = [5 10] X1 + [1 2] X2 + [4 8] X3 + [1]
[0 0] [0 0] [0 0] [8]
>= [5 2] X1 + [1 2] X2 + [4 0] X3 + [1]
[0 0] [0 0] [0 0] [4]
= [ok(f(X1, X2, X3))]
[proper(f(X1, X2, X3))] = [ 5 0] X1 + [1 0] X2 + [ 4 0] X3 + [1]
[15 0] [3 0] [12 0] [6]
>= [5 0] X1 + [1 0] X2 + [4 0] X3 + [1]
[3 0] [3 0] [0 0] [6]
= [f(proper(X1), proper(X2), proper(X3))]
[proper(a())] = [1]
[6]
>= [1]
[4]
= [ok(a())]
[proper(b())] = [1]
[6]
>= [1]
[4]
= [ok(b())]
[active^#(f(X1, X2, X3))] = [5 6] X1 + [1 6] X2 + [4 0] X3 + [1]
[0 0] [0 0] [0 0] [0]
> [3 2] X1 + [1 6] X2 + [0]
[0 0] [0 0] [0]
= [c_1(f^#(X1, active(X2), X3), active^#(X2))]
[f^#(X1, mark(X2), X3)] = [1 1] X1 + [0]
[1 0] [0]
>= [1 1] X1 + [0]
[0 0] [0]
= [c_4(f^#(X1, X2, X3))]
[f^#(ok(X1), ok(X2), ok(X3))] = [1 2] X1 + [4]
[1 2] [0]
> [1 1] X1 + [3]
[0 0] [0]
= [c_5(f^#(X1, X2, X3))]
[proper^#(f(X1, X2, X3))] = [15 0] X1 + [3 0] X2 + [12 0] X3 + [3]
[ 5 0] [1 0] [ 4 0] [1]
>= [14 0] X1 + [3 0] X2 + [4 0] X3 + [3]
[ 0 0] [0 0] [0 0] [0]
= [c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3))]
[top^#(mark(X))] = [3 0] X + [0]
[4 0] [4]
>= [3 0] X + [0]
[1 0] [0]
= [proper^#(X)]
[top^#(mark(X))] = [3 0] X + [0]
[4 0] [4]
>= [3 0] X + [0]
[4 0] [4]
= [top^#(proper(X))]
[top^#(ok(X))] = [3 6] X + [0]
[4 8] [4]
>= [1 6] X + [0]
[0 0] [0]
= [active^#(X)]
[top^#(ok(X))] = [3 6] X + [0]
[4 8] [4]
>= [3 0] X + [0]
[4 0] [4]
= [top^#(active(X))]
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:
{ proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3)) }
Weak DPs:
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, top^#(mark(X)) -> proper^#(X)
, top^#(mark(X)) -> top^#(proper(X))
, top^#(ok(X)) -> active^#(X)
, top^#(ok(X)) -> top^#(active(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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.
{ active^#(f(X1, X2, X3)) ->
c_1(f^#(X1, active(X2), X3), active^#(X2))
, f^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, top^#(ok(X)) -> active^#(X) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3)) }
Weak DPs:
{ top^#(mark(X)) -> proper^#(X)
, top^#(mark(X)) -> top^#(proper(X))
, top^#(ok(X)) -> top^#(active(X)) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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:
{ proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ proper^#(f(X1, X2, X3)) ->
c_1(proper^#(X1), proper^#(X2), proper^#(X3)) }
Weak DPs:
{ top^#(mark(X)) -> c_2(proper^#(X))
, top^#(mark(X)) -> c_3(top^#(proper(X)))
, top^#(ok(X)) -> c_4(top^#(active(X))) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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: proper^#(f(X1, X2, X3)) ->
c_1(proper^#(X1), proper^#(X2), proper^#(X3))
, 2: top^#(mark(X)) -> c_2(proper^#(X)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1, 2, 3}, Uargs(c_2) = {1}, Uargs(c_3) = {1},
Uargs(c_4) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[active](x1) = [1] x1 + [0]
[f](x1, x2, x3) = [1] x1 + [2] x2 + [1] x3 + [1]
[a] = [0]
[mark](x1) = [1] x1 + [0]
[b] = [0]
[proper](x1) = [1] x1 + [0]
[ok](x1) = [1] x1 + [0]
[active^#](x1) = [0]
[c_1](x1, x2) = [0]
[f^#](x1, x2, x3) = [0]
[c_4](x1) = [0]
[c_5](x1) = [0]
[proper^#](x1) = [5] x1 + [0]
[c_6](x1, x2, x3, x4) = [0]
[top^#](x1) = [7] x1 + [2]
[c] = [0]
[c_1](x1, x2, x3) = [1] x1 + [2] x2 + [1] x3 + [0]
[c_2](x1) = [1] x1 + [1]
[c_3](x1) = [1] x1 + [0]
[c_4](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[active(f(X1, X2, X3))] = [1] X1 + [2] X2 + [1] X3 + [1]
>= [1] X1 + [2] X2 + [1] X3 + [1]
= [f(X1, active(X2), X3)]
[active(f(a(), X, X))] = [3] X + [1]
>= [1] X + [1]
= [mark(f(X, b(), b()))]
[active(b())] = [0]
>= [0]
= [mark(a())]
[f(X1, mark(X2), X3)] = [1] X1 + [2] X2 + [1] X3 + [1]
>= [1] X1 + [2] X2 + [1] X3 + [1]
= [mark(f(X1, X2, X3))]
[f(ok(X1), ok(X2), ok(X3))] = [1] X1 + [2] X2 + [1] X3 + [1]
>= [1] X1 + [2] X2 + [1] X3 + [1]
= [ok(f(X1, X2, X3))]
[proper(f(X1, X2, X3))] = [1] X1 + [2] X2 + [1] X3 + [1]
>= [1] X1 + [2] X2 + [1] X3 + [1]
= [f(proper(X1), proper(X2), proper(X3))]
[proper(a())] = [0]
>= [0]
= [ok(a())]
[proper(b())] = [0]
>= [0]
= [ok(b())]
[proper^#(f(X1, X2, X3))] = [5] X1 + [10] X2 + [5] X3 + [5]
> [5] X1 + [10] X2 + [5] X3 + [0]
= [c_1(proper^#(X1), proper^#(X2), proper^#(X3))]
[top^#(mark(X))] = [7] X + [2]
> [5] X + [1]
= [c_2(proper^#(X))]
[top^#(mark(X))] = [7] X + [2]
>= [7] X + [2]
= [c_3(top^#(proper(X)))]
[top^#(ok(X))] = [7] X + [2]
>= [7] X + [2]
= [c_4(top^#(active(X)))]
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:
{ proper^#(f(X1, X2, X3)) ->
c_1(proper^#(X1), proper^#(X2), proper^#(X3))
, top^#(mark(X)) -> c_2(proper^#(X))
, top^#(mark(X)) -> c_3(top^#(proper(X)))
, top^#(ok(X)) -> c_4(top^#(active(X))) }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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.
{ proper^#(f(X1, X2, X3)) ->
c_1(proper^#(X1), proper^#(X2), proper^#(X3))
, top^#(mark(X)) -> c_2(proper^#(X))
, top^#(mark(X)) -> c_3(top^#(proper(X)))
, top^#(ok(X)) -> c_4(top^#(active(X))) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(1)).
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(a(), X, X)) -> mark(f(X, b(), b()))
, active(b()) -> mark(a())
, f(X1, mark(X2), X3) -> mark(f(X1, X2, X3))
, f(ok(X1), ok(X2), ok(X3)) -> ok(f(X1, X2, X3))
, proper(f(X1, X2, X3)) -> f(proper(X1), proper(X2), proper(X3))
, proper(a()) -> ok(a())
, proper(b()) -> ok(b()) }
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))