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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c())
, 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(b(), X, c())) -> c_2(f^#(X, c(), X))
, active^#(c()) -> 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^#(b()) -> c_7()
, proper^#(c()) -> 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(b(), X, c())) -> c_2(f^#(X, c(), X))
, active^#(c()) -> 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^#(b()) -> c_7()
, proper^#(c()) -> 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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c())
, 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(b(), X, c())) -> c_2(f^#(X, c(), X))
, 3: active^#(c()) -> 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^#(b()) -> c_7()
, 8: proper^#(c()) -> 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(b(), X, c())) -> c_2(f^#(X, c(), X))
, active^#(c()) -> c_3()
, proper^#(b()) -> c_7()
, proper^#(c()) -> c_8() }
Weak Trs:
{ active(f(X1, X2, X3)) -> f(X1, active(X2), X3)
, active(f(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c())
, 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(b(), X, c())) -> c_2(f^#(X, c(), X))
, active^#(c()) -> c_3()
, proper^#(b()) -> c_7()
, proper^#(c()) -> 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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c())
, 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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
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)) }
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] [0]
[f](x1, x2, x3) = [0 0] x1 + [3 0] x2 + [0 0] x3 + [0]
[4 1] [1 4] [0 3] [0]
[b] = [1]
[1]
[c] = [0]
[3]
[mark](x1) = [0 0] x1 + [0]
[1 1] [1]
[proper](x1) = [1 0] x1 + [0]
[0 1] [0]
[ok](x1) = [1 0] x1 + [0]
[0 1] [0]
[active^#](x1) = [0]
[0]
[proper^#](x1) = [0]
[1]
[top^#](x1) = [4 4] x1 + [2]
[1 0] [0]
[c_9](x1, x2) = [1 0] x1 + [1 0] x2 + [0]
[0 0] [0 0] [0]
[c_10](x1, x2) = [1 1] x1 + [1 1] 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]
[4 1] [4 4] [0 3] [0]
>= [0 0] X1 + [0 0] X2 + [0 0] X3 + [0]
[4 1] [4 4] [0 3] [0]
= [f(X1, active(X2), X3)]
[active(f(b(), X, c()))] = [0 0] X + [0]
[4 4] [14]
>= [0 0] X + [0]
[4 4] [13]
= [mark(f(X, c(), X))]
[active(c())] = [0]
[3]
>= [0]
[3]
= [mark(b())]
[f(X1, mark(X2), X3)] = [0 0] X1 + [0 0] X2 + [0 0] X3 + [0]
[4 1] [4 4] [0 3] [4]
>= [0 0] X1 + [0 0] X2 + [0 0] X3 + [0]
[4 1] [4 4] [0 3] [1]
= [mark(f(X1, X2, X3))]
[f(ok(X1), ok(X2), ok(X3))] = [0 0] X1 + [3 0] X2 + [0 0] X3 + [0]
[4 1] [1 4] [0 3] [0]
>= [0 0] X1 + [3 0] X2 + [0 0] X3 + [0]
[4 1] [1 4] [0 3] [0]
= [ok(f(X1, X2, X3))]
[proper(f(X1, X2, X3))] = [0 0] X1 + [3 0] X2 + [0 0] X3 + [0]
[4 1] [1 4] [0 3] [0]
>= [0 0] X1 + [3 0] X2 + [0 0] X3 + [0]
[4 1] [1 4] [0 3] [0]
= [f(proper(X1), proper(X2), proper(X3))]
[proper(b())] = [1]
[1]
>= [1]
[1]
= [ok(b())]
[proper(c())] = [0]
[3]
>= [0]
[3]
= [ok(c())]
[top^#(mark(X))] = [4 4] X + [6]
[0 0] [0]
> [4 4] X + [2]
[0 0] [0]
= [c_9(top^#(proper(X)), proper^#(X))]
[top^#(ok(X))] = [4 4] X + [2]
[1 0] [0]
>= [4 4] X + [2]
[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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We use the processor 'matrix interpretation of dimension 3' to
orient following rules strictly.
DPs:
{ 1: top^#(ok(X)) -> c_10(top^#(active(X)), active^#(X)) }
Trs:
{ active(f(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b()) }
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)).
[0 0 1] [0]
[active](x1) = [0 0 1] x1 + [0]
[1 0 0] [0]
[2 0 1] [2 0 2] [0 0 1] [0]
[f](x1, x2, x3) = [2 0 1] x1 + [1 1 2] x2 + [0 0 1] x3 + [0]
[2 0 1] [2 0 2] [0 0 1] [0]
[2]
[b] = [1]
[0]
[0]
[c] = [2]
[3]
[1 0 0] [0]
[mark](x1) = [1 0 0] x1 + [1]
[0 0 1] [0]
[1 0 0] [0]
[proper](x1) = [1 0 0] x1 + [1]
[0 0 1] [0]
[1 0 0] [0]
[ok](x1) = [1 0 0] x1 + [1]
[0 0 1] [0]
[3]
[active^#](x1) = [3]
[1]
[0]
[proper^#](x1) = [0]
[0]
[2 5 7] [4]
[top^#](x1) = [0 0 0] x1 + [2]
[0 0 0] [0]
[1 0 0] [0]
[c_9](x1, x2) = [0 0 0] x1 + [0]
[0 0 0] [0]
[1 2 0] [0]
[c_10](x1, x2) = [0 0 0] x1 + [0]
[0 0 0] [0]
The order satisfies the following ordering constraints:
[active(f(X1, X2, X3))] = [2 0 1] [2 0 2] [0 0 1] [0]
[2 0 1] X1 + [2 0 2] X2 + [0 0 1] X3 + [0]
[2 0 1] [2 0 2] [0 0 1] [0]
>= [2 0 1] [2 0 2] [0 0 1] [0]
[2 0 1] X1 + [2 0 2] X2 + [0 0 1] X3 + [0]
[2 0 1] [2 0 2] [0 0 1] [0]
= [f(X1, active(X2), X3)]
[active(f(b(), X, c()))] = [2 0 2] [7]
[2 0 2] X + [7]
[2 0 2] [7]
> [2 0 2] [6]
[2 0 2] X + [7]
[2 0 2] [6]
= [mark(f(X, c(), X))]
[active(c())] = [3]
[3]
[0]
> [2]
[3]
[0]
= [mark(b())]
[f(X1, mark(X2), X3)] = [2 0 1] [2 0 2] [0 0 1] [0]
[2 0 1] X1 + [2 0 2] X2 + [0 0 1] X3 + [1]
[2 0 1] [2 0 2] [0 0 1] [0]
>= [2 0 1] [2 0 2] [0 0 1] [0]
[2 0 1] X1 + [2 0 2] X2 + [0 0 1] X3 + [1]
[2 0 1] [2 0 2] [0 0 1] [0]
= [mark(f(X1, X2, X3))]
[f(ok(X1), ok(X2), ok(X3))] = [2 0 1] [2 0 2] [0 0 1] [0]
[2 0 1] X1 + [2 0 2] X2 + [0 0 1] X3 + [1]
[2 0 1] [2 0 2] [0 0 1] [0]
>= [2 0 1] [2 0 2] [0 0 1] [0]
[2 0 1] X1 + [2 0 2] X2 + [0 0 1] X3 + [1]
[2 0 1] [2 0 2] [0 0 1] [0]
= [ok(f(X1, X2, X3))]
[proper(f(X1, X2, X3))] = [2 0 1] [2 0 2] [0 0 1] [0]
[2 0 1] X1 + [2 0 2] X2 + [0 0 1] X3 + [1]
[2 0 1] [2 0 2] [0 0 1] [0]
>= [2 0 1] [2 0 2] [0 0 1] [0]
[2 0 1] X1 + [2 0 2] X2 + [0 0 1] X3 + [1]
[2 0 1] [2 0 2] [0 0 1] [0]
= [f(proper(X1), proper(X2), proper(X3))]
[proper(b())] = [2]
[3]
[0]
>= [2]
[3]
[0]
= [ok(b())]
[proper(c())] = [0]
[1]
[3]
>= [0]
[1]
[3]
= [ok(c())]
[top^#(mark(X))] = [7 0 7] [9]
[0 0 0] X + [2]
[0 0 0] [0]
>= [7 0 7] [9]
[0 0 0] X + [0]
[0 0 0] [0]
= [c_9(top^#(proper(X)), proper^#(X))]
[top^#(ok(X))] = [7 0 7] [9]
[0 0 0] X + [2]
[0 0 0] [0]
> [7 0 7] [8]
[0 0 0] X + [0]
[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(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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
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:
{ 3: f^#(ok(X1), ok(X2), ok(X3)) -> c_5(f^#(X1, X2, X3))
, 5: top^#(mark(X)) -> proper^#(X)
, 7: top^#(ok(X)) -> active^#(X) }
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]
[2 0] [1]
[f](x1, x2, x3) = [2 0] x1 + [4 0] x2 + [2 0] x3 + [3]
[0 0] [0 1] [0 0] [0]
[b] = [0]
[0]
[c] = [0]
[0]
[mark](x1) = [1 1] x1 + [0]
[0 0] [0]
[proper](x1) = [1 0] x1 + [0]
[2 0] [2]
[ok](x1) = [1 2] x1 + [0]
[0 0] [2]
[active^#](x1) = [1 2] x1 + [5]
[0 0] [1]
[c_1](x1, x2) = [1 0] x1 + [1 2] x2 + [0]
[0 0] [0 0] [0]
[f^#](x1, x2, x3) = [0 0] x1 + [1 1] x2 + [1 0] x3 + [0]
[1 0] [0 0] [0 0] [1]
[c_4](x1) = [1 0] x1 + [0]
[0 0] [0]
[c_5](x1) = [1 0] x1 + [1]
[0 0] [0]
[proper^#](x1) = [1 0] x1 + [0]
[1 0] [2]
[c_6](x1, x2, x3, x4) = [1 1] x1 + [1 0] x2 + [1 0] x3 + [1
0] x4 + [0]
[0 0] [0 0] [0 0] [0
0] [0]
[top^#](x1) = [1 0] x1 + [7]
[4 0] [4]
The order satisfies the following ordering constraints:
[active(f(X1, X2, X3))] = [2 0] X1 + [4 0] X2 + [2 0] X3 + [3]
[4 0] [8 0] [4 0] [7]
>= [2 0] X1 + [4 0] X2 + [2 0] X3 + [3]
[0 0] [2 0] [0 0] [1]
= [f(X1, active(X2), X3)]
[active(f(b(), X, c()))] = [4 0] X + [3]
[8 0] [7]
>= [4 0] X + [3]
[0 0] [0]
= [mark(f(X, c(), X))]
[active(c())] = [0]
[1]
>= [0]
[0]
= [mark(b())]
[f(X1, mark(X2), X3)] = [2 0] X1 + [4 4] X2 + [2 0] X3 + [3]
[0 0] [0 0] [0 0] [0]
>= [2 0] X1 + [4 1] X2 + [2 0] X3 + [3]
[0 0] [0 0] [0 0] [0]
= [mark(f(X1, X2, X3))]
[f(ok(X1), ok(X2), ok(X3))] = [2 4] X1 + [4 8] X2 + [2 4] X3 + [3]
[0 0] [0 0] [0 0] [2]
>= [2 0] X1 + [4 2] X2 + [2 0] X3 + [3]
[0 0] [0 0] [0 0] [2]
= [ok(f(X1, X2, X3))]
[proper(f(X1, X2, X3))] = [2 0] X1 + [4 0] X2 + [2 0] X3 + [3]
[4 0] [8 0] [4 0] [8]
>= [2 0] X1 + [4 0] X2 + [2 0] X3 + [3]
[0 0] [2 0] [0 0] [2]
= [f(proper(X1), proper(X2), proper(X3))]
[proper(b())] = [0]
[2]
>= [0]
[2]
= [ok(b())]
[proper(c())] = [0]
[2]
>= [0]
[2]
= [ok(c())]
[active^#(f(X1, X2, X3))] = [2 0] X1 + [4 2] X2 + [2 0] X3 + [8]
[0 0] [0 0] [0 0] [1]
>= [4 2] X2 + [1 0] X3 + [8]
[0 0] [0 0] [0]
= [c_1(f^#(X1, active(X2), X3), active^#(X2))]
[f^#(X1, mark(X2), X3)] = [0 0] X1 + [1 1] X2 + [1 0] X3 + [0]
[1 0] [0 0] [0 0] [1]
>= [1 1] X2 + [1 0] X3 + [0]
[0 0] [0 0] [0]
= [c_4(f^#(X1, X2, X3))]
[f^#(ok(X1), ok(X2), ok(X3))] = [0 0] X1 + [1 2] X2 + [1 2] X3 + [2]
[1 2] [0 0] [0 0] [1]
> [1 1] X2 + [1 0] X3 + [1]
[0 0] [0 0] [0]
= [c_5(f^#(X1, X2, X3))]
[proper^#(f(X1, X2, X3))] = [2 0] X1 + [4 0] X2 + [2 0] X3 + [3]
[2 0] [4 0] [2 0] [5]
>= [2 0] X1 + [4 0] X2 + [2 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))] = [1 1] X + [7]
[4 4] [4]
> [1 0] X + [0]
[1 0] [2]
= [proper^#(X)]
[top^#(mark(X))] = [1 1] X + [7]
[4 4] [4]
>= [1 0] X + [7]
[4 0] [4]
= [top^#(proper(X))]
[top^#(ok(X))] = [1 2] X + [7]
[4 8] [4]
> [1 2] X + [5]
[0 0] [1]
= [active^#(X)]
[top^#(ok(X))] = [1 2] X + [7]
[4 8] [4]
>= [1 0] X + [7]
[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:
{ 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))
, proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3)) }
Weak DPs:
{ 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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
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^#(X1, mark(X2), X3) -> c_4(f^#(X1, X2, X3))
, 3: proper^#(f(X1, X2, X3)) ->
c_6(f^#(proper(X1), proper(X2), proper(X3)),
proper^#(X1),
proper^#(X2),
proper^#(X3))
, 5: top^#(mark(X)) -> proper^#(X) }
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 6] x1 + [0]
[1 0] [2]
[f](x1, x2, x3) = [1 5] x1 + [2 0] x2 + [1 4] x3 + [7]
[0 0] [0 2] [0 0] [0]
[b] = [2]
[0]
[c] = [2]
[0]
[mark](x1) = [1 3] x1 + [0]
[0 0] [1]
[proper](x1) = [1 3] x1 + [0]
[0 0] [0]
[ok](x1) = [1 6] x1 + [0]
[0 0] [0]
[active^#](x1) = [2 7] x1 + [1]
[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 + [4]
[1 0] [0 1] [1]
[c_4](x1) = [1 0] x1 + [0]
[0 0] [0]
[c_5](x1) = [1 0] x1 + [0]
[0 0] [0]
[proper^#](x1) = [1 3] x1 + [0]
[0 0] [0]
[c_6](x1, x2, x3, x4) = [1 0] x1 + [1 0] x2 + [1 1] x3 + [1
0] x4 + [0]
[0 0] [0 0] [0 0] [0
0] [0]
[top^#](x1) = [2 0] x1 + [1]
[2 0] [4]
The order satisfies the following ordering constraints:
[active(f(X1, X2, X3))] = [1 5] X1 + [2 12] X2 + [1 4] X3 + [7]
[1 5] [2 0] [1 4] [9]
>= [1 5] X1 + [2 12] X2 + [1 4] X3 + [7]
[0 0] [2 0] [0 0] [4]
= [f(X1, active(X2), X3)]
[active(f(b(), X, c()))] = [2 12] X + [11]
[2 0] [13]
>= [2 9] X + [11]
[0 0] [1]
= [mark(f(X, c(), X))]
[active(c())] = [2]
[4]
>= [2]
[1]
= [mark(b())]
[f(X1, mark(X2), X3)] = [1 5] X1 + [2 6] X2 + [1 4] X3 + [7]
[0 0] [0 0] [0 0] [2]
>= [1 5] X1 + [2 6] X2 + [1 4] X3 + [7]
[0 0] [0 0] [0 0] [1]
= [mark(f(X1, X2, X3))]
[f(ok(X1), ok(X2), ok(X3))] = [1 6] X1 + [2 12] X2 + [1 6] X3 + [7]
[0 0] [0 0] [0 0] [0]
>= [1 5] X1 + [2 12] X2 + [1 4] X3 + [7]
[0 0] [0 0] [0 0] [0]
= [ok(f(X1, X2, X3))]
[proper(f(X1, X2, X3))] = [1 5] X1 + [2 6] X2 + [1 4] X3 + [7]
[0 0] [0 0] [0 0] [0]
>= [1 3] X1 + [2 6] X2 + [1 3] X3 + [7]
[0 0] [0 0] [0 0] [0]
= [f(proper(X1), proper(X2), proper(X3))]
[proper(b())] = [2]
[0]
>= [2]
[0]
= [ok(b())]
[proper(c())] = [2]
[0]
>= [2]
[0]
= [ok(c())]
[active^#(f(X1, X2, X3))] = [2 10] X1 + [4 14] X2 + [2 8] X3 + [15]
[0 0] [0 2] [0 0] [0]
> [4 14] X2 + [7]
[0 0] [0]
= [c_1(f^#(X1, active(X2), X3), active^#(X2))]
[f^#(X1, mark(X2), X3)] = [1 3] X2 + [0 0] X3 + [5]
[1 3] [0 1] [1]
> [1 1] X2 + [4]
[0 0] [0]
= [c_4(f^#(X1, X2, X3))]
[f^#(ok(X1), ok(X2), ok(X3))] = [1 6] X2 + [4]
[1 6] [1]
>= [1 1] X2 + [4]
[0 0] [0]
= [c_5(f^#(X1, X2, X3))]
[proper^#(f(X1, X2, X3))] = [1 5] X1 + [2 6] X2 + [1 4] X3 + [7]
[0 0] [0 0] [0 0] [0]
> [1 3] X1 + [2 6] X2 + [1 3] X3 + [4]
[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))] = [2 6] X + [1]
[2 6] [4]
> [1 3] X + [0]
[0 0] [0]
= [proper^#(X)]
[top^#(mark(X))] = [2 6] X + [1]
[2 6] [4]
>= [2 6] X + [1]
[2 6] [4]
= [top^#(proper(X))]
[top^#(ok(X))] = [2 12] X + [1]
[2 12] [4]
>= [2 7] X + [1]
[0 1] [0]
= [active^#(X)]
[top^#(ok(X))] = [2 12] X + [1]
[2 12] [4]
>= [2 12] X + [1]
[2 12] [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:
{ 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)) -> 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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
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.
{ 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)) -> proper^#(X)
, top^#(mark(X)) -> top^#(proper(X))
, top^#(ok(X)) -> active^#(X)
, top^#(ok(X)) -> 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(b(), X, c())) -> mark(f(X, c(), X))
, active(c()) -> mark(b())
, 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(b()) -> ok(b())
, proper(c()) -> ok(c()) }
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))