We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ circ(s, id()) -> s
, circ(circ(s, t), u) -> circ(s, circ(t, u))
, circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
, circ(cons(lift(), s), circ(cons(lift(), t), u)) ->
circ(cons(lift(), circ(s, t)), u)
, circ(cons(lift(), s), cons(a, t)) -> cons(a, circ(s, t))
, circ(cons(lift(), s), cons(lift(), t)) ->
cons(lift(), circ(s, t))
, circ(id(), s) -> s
, msubst(a, id()) -> a
, msubst(msubst(a, s), t) -> msubst(a, circ(s, t))
, subst(a, id()) -> a }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
Arguments of following rules are not normal-forms:
{ circ(cons(lift(), s), circ(cons(lift(), t), u)) ->
circ(cons(lift(), circ(s, t)), u) }
All above mentioned rules can be savely removed.
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict Trs:
{ circ(s, id()) -> s
, circ(circ(s, t), u) -> circ(s, circ(t, u))
, circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
, circ(cons(lift(), s), cons(a, t)) -> cons(a, circ(s, t))
, circ(cons(lift(), s), cons(lift(), t)) ->
cons(lift(), circ(s, t))
, circ(id(), s) -> s
, msubst(a, id()) -> a
, msubst(msubst(a, s), t) -> msubst(a, circ(s, t))
, subst(a, id()) -> a }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We add the following dependency tuples:
Strict DPs:
{ circ^#(s, id()) -> c_1()
, circ^#(circ(s, t), u) -> c_2(circ^#(s, circ(t, u)), circ^#(t, u))
, circ^#(cons(a, s), t) -> c_3(msubst^#(a, t), circ^#(s, t))
, circ^#(cons(lift(), s), cons(a, t)) -> c_4(circ^#(s, t))
, circ^#(cons(lift(), s), cons(lift(), t)) -> c_5(circ^#(s, t))
, circ^#(id(), s) -> c_6()
, msubst^#(a, id()) -> c_7()
, msubst^#(msubst(a, s), t) ->
c_8(msubst^#(a, circ(s, t)), circ^#(s, t))
, subst^#(a, id()) -> c_9() }
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:
{ circ^#(s, id()) -> c_1()
, circ^#(circ(s, t), u) -> c_2(circ^#(s, circ(t, u)), circ^#(t, u))
, circ^#(cons(a, s), t) -> c_3(msubst^#(a, t), circ^#(s, t))
, circ^#(cons(lift(), s), cons(a, t)) -> c_4(circ^#(s, t))
, circ^#(cons(lift(), s), cons(lift(), t)) -> c_5(circ^#(s, t))
, circ^#(id(), s) -> c_6()
, msubst^#(a, id()) -> c_7()
, msubst^#(msubst(a, s), t) ->
c_8(msubst^#(a, circ(s, t)), circ^#(s, t))
, subst^#(a, id()) -> c_9() }
Weak Trs:
{ circ(s, id()) -> s
, circ(circ(s, t), u) -> circ(s, circ(t, u))
, circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
, circ(cons(lift(), s), cons(a, t)) -> cons(a, circ(s, t))
, circ(cons(lift(), s), cons(lift(), t)) ->
cons(lift(), circ(s, t))
, circ(id(), s) -> s
, msubst(a, id()) -> a
, msubst(msubst(a, s), t) -> msubst(a, circ(s, t))
, subst(a, id()) -> a }
Obligation:
innermost runtime complexity
Answer:
YES(O(1),O(n^1))
We estimate the number of application of {1,6,7,9} by applications
of Pre({1,6,7,9}) = {2,3,4,5,8}. Here rules are labeled as follows:
DPs:
{ 1: circ^#(s, id()) -> c_1()
, 2: circ^#(circ(s, t), u) ->
c_2(circ^#(s, circ(t, u)), circ^#(t, u))
, 3: circ^#(cons(a, s), t) -> c_3(msubst^#(a, t), circ^#(s, t))
, 4: circ^#(cons(lift(), s), cons(a, t)) -> c_4(circ^#(s, t))
, 5: circ^#(cons(lift(), s), cons(lift(), t)) -> c_5(circ^#(s, t))
, 6: circ^#(id(), s) -> c_6()
, 7: msubst^#(a, id()) -> c_7()
, 8: msubst^#(msubst(a, s), t) ->
c_8(msubst^#(a, circ(s, t)), circ^#(s, t))
, 9: subst^#(a, id()) -> c_9() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ circ^#(circ(s, t), u) -> c_2(circ^#(s, circ(t, u)), circ^#(t, u))
, circ^#(cons(a, s), t) -> c_3(msubst^#(a, t), circ^#(s, t))
, circ^#(cons(lift(), s), cons(a, t)) -> c_4(circ^#(s, t))
, circ^#(cons(lift(), s), cons(lift(), t)) -> c_5(circ^#(s, t))
, msubst^#(msubst(a, s), t) ->
c_8(msubst^#(a, circ(s, t)), circ^#(s, t)) }
Weak DPs:
{ circ^#(s, id()) -> c_1()
, circ^#(id(), s) -> c_6()
, msubst^#(a, id()) -> c_7()
, subst^#(a, id()) -> c_9() }
Weak Trs:
{ circ(s, id()) -> s
, circ(circ(s, t), u) -> circ(s, circ(t, u))
, circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
, circ(cons(lift(), s), cons(a, t)) -> cons(a, circ(s, t))
, circ(cons(lift(), s), cons(lift(), t)) ->
cons(lift(), circ(s, t))
, circ(id(), s) -> s
, msubst(a, id()) -> a
, msubst(msubst(a, s), t) -> msubst(a, circ(s, t))
, subst(a, id()) -> a }
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.
{ circ^#(s, id()) -> c_1()
, circ^#(id(), s) -> c_6()
, msubst^#(a, id()) -> c_7()
, subst^#(a, id()) -> c_9() }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ circ^#(circ(s, t), u) -> c_2(circ^#(s, circ(t, u)), circ^#(t, u))
, circ^#(cons(a, s), t) -> c_3(msubst^#(a, t), circ^#(s, t))
, circ^#(cons(lift(), s), cons(a, t)) -> c_4(circ^#(s, t))
, circ^#(cons(lift(), s), cons(lift(), t)) -> c_5(circ^#(s, t))
, msubst^#(msubst(a, s), t) ->
c_8(msubst^#(a, circ(s, t)), circ^#(s, t)) }
Weak Trs:
{ circ(s, id()) -> s
, circ(circ(s, t), u) -> circ(s, circ(t, u))
, circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
, circ(cons(lift(), s), cons(a, t)) -> cons(a, circ(s, t))
, circ(cons(lift(), s), cons(lift(), t)) ->
cons(lift(), circ(s, t))
, circ(id(), s) -> s
, msubst(a, id()) -> a
, msubst(msubst(a, s), t) -> msubst(a, circ(s, t))
, subst(a, id()) -> a }
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:
{ circ^#(circ(s, t), u) -> c_2(circ^#(s, circ(t, u)), circ^#(t, u))
, msubst^#(msubst(a, s), t) ->
c_8(msubst^#(a, circ(s, t)), circ^#(s, t)) }
We are left with following problem, upon which TcT provides the
certificate YES(O(1),O(n^1)).
Strict DPs:
{ circ^#(circ(s, t), u) -> c_1(circ^#(t, u))
, circ^#(cons(a, s), t) -> c_2(msubst^#(a, t), circ^#(s, t))
, circ^#(cons(lift(), s), cons(a, t)) -> c_3(circ^#(s, t))
, circ^#(cons(lift(), s), cons(lift(), t)) -> c_4(circ^#(s, t))
, msubst^#(msubst(a, s), t) -> c_5(circ^#(s, t)) }
Weak Trs:
{ circ(s, id()) -> s
, circ(circ(s, t), u) -> circ(s, circ(t, u))
, circ(cons(a, s), t) -> cons(msubst(a, t), circ(s, t))
, circ(cons(lift(), s), cons(a, t)) -> cons(a, circ(s, t))
, circ(cons(lift(), s), cons(lift(), t)) ->
cons(lift(), circ(s, t))
, circ(id(), s) -> s
, msubst(a, id()) -> a
, msubst(msubst(a, s), t) -> msubst(a, circ(s, t))
, subst(a, id()) -> a }
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:
{ circ^#(circ(s, t), u) -> c_1(circ^#(t, u))
, circ^#(cons(a, s), t) -> c_2(msubst^#(a, t), circ^#(s, t))
, circ^#(cons(lift(), s), cons(a, t)) -> c_3(circ^#(s, t))
, circ^#(cons(lift(), s), cons(lift(), t)) -> c_4(circ^#(s, t))
, msubst^#(msubst(a, s), t) -> c_5(circ^#(s, t)) }
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: circ^#(circ(s, t), u) -> c_1(circ^#(t, u)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_3) = {1},
Uargs(c_4) = {1}, Uargs(c_5) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[circ](x1, x2) = [2] x2 + [2]
[cons](x1, x2) = [1] x1 + [1] x2 + [0]
[msubst](x1, x2) = [1] x1 + [5] x2 + [0]
[lift] = [0]
[circ^#](x1, x2) = [4] x1 + [0]
[msubst^#](x1, x2) = [1] x1 + [0]
[c_1](x1) = [1] x1 + [1]
[c_2](x1, x2) = [4] x1 + [1] x2 + [0]
[c_3](x1) = [1] x1 + [0]
[c_4](x1) = [1] x1 + [0]
[c_5](x1) = [1] x1 + [0]
The order satisfies the following ordering constraints:
[circ^#(circ(s, t), u)] = [8] t + [8]
> [4] t + [1]
= [c_1(circ^#(t, u))]
[circ^#(cons(a, s), t)] = [4] a + [4] s + [0]
>= [4] a + [4] s + [0]
= [c_2(msubst^#(a, t), circ^#(s, t))]
[circ^#(cons(lift(), s), cons(a, t))] = [4] s + [0]
>= [4] s + [0]
= [c_3(circ^#(s, t))]
[circ^#(cons(lift(), s), cons(lift(), t))] = [4] s + [0]
>= [4] s + [0]
= [c_4(circ^#(s, t))]
[msubst^#(msubst(a, s), t)] = [1] a + [5] s + [0]
>= [4] s + [0]
= [c_5(circ^#(s, t))]
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:
{ circ^#(cons(a, s), t) -> c_2(msubst^#(a, t), circ^#(s, t))
, circ^#(cons(lift(), s), cons(a, t)) -> c_3(circ^#(s, t))
, circ^#(cons(lift(), s), cons(lift(), t)) -> c_4(circ^#(s, t))
, msubst^#(msubst(a, s), t) -> c_5(circ^#(s, t)) }
Weak DPs: { circ^#(circ(s, t), u) -> c_1(circ^#(t, u)) }
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: circ^#(cons(a, s), t) -> c_2(msubst^#(a, t), circ^#(s, t))
, 2: circ^#(cons(lift(), s), cons(a, t)) -> c_3(circ^#(s, t))
, 3: circ^#(cons(lift(), s), cons(lift(), t)) -> c_4(circ^#(s, t))
, 4: msubst^#(msubst(a, s), t) -> c_5(circ^#(s, t)) }
Sub-proof:
----------
The following argument positions are usable:
Uargs(c_1) = {1}, Uargs(c_2) = {1, 2}, Uargs(c_3) = {1},
Uargs(c_4) = {1}, Uargs(c_5) = {1}
TcT has computed the following constructor-based matrix
interpretation satisfying not(EDA).
[circ](x1, x2) = [4] x1 + [2] x2 + [0]
[cons](x1, x2) = [1] x1 + [1] x2 + [4]
[msubst](x1, x2) = [1] x1 + [4] x2 + [0]
[lift] = [0]
[circ^#](x1, x2) = [2] x1 + [0]
[msubst^#](x1, x2) = [2] x1 + [1]
[c_1](x1) = [2] x1 + [0]
[c_2](x1, x2) = [1] x1 + [1] x2 + [6]
[c_3](x1) = [1] x1 + [0]
[c_4](x1) = [1] x1 + [0]
[c_5](x1) = [4] x1 + [0]
The order satisfies the following ordering constraints:
[circ^#(circ(s, t), u)] = [8] s + [4] t + [0]
>= [4] t + [0]
= [c_1(circ^#(t, u))]
[circ^#(cons(a, s), t)] = [2] a + [2] s + [8]
> [2] a + [2] s + [7]
= [c_2(msubst^#(a, t), circ^#(s, t))]
[circ^#(cons(lift(), s), cons(a, t))] = [2] s + [8]
> [2] s + [0]
= [c_3(circ^#(s, t))]
[circ^#(cons(lift(), s), cons(lift(), t))] = [2] s + [8]
> [2] s + [0]
= [c_4(circ^#(s, t))]
[msubst^#(msubst(a, s), t)] = [2] a + [8] s + [1]
> [8] s + [0]
= [c_5(circ^#(s, t))]
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:
{ circ^#(circ(s, t), u) -> c_1(circ^#(t, u))
, circ^#(cons(a, s), t) -> c_2(msubst^#(a, t), circ^#(s, t))
, circ^#(cons(lift(), s), cons(a, t)) -> c_3(circ^#(s, t))
, circ^#(cons(lift(), s), cons(lift(), t)) -> c_4(circ^#(s, t))
, msubst^#(msubst(a, s), t) -> c_5(circ^#(s, t)) }
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.
{ circ^#(circ(s, t), u) -> c_1(circ^#(t, u))
, circ^#(cons(a, s), t) -> c_2(msubst^#(a, t), circ^#(s, t))
, circ^#(cons(lift(), s), cons(a, t)) -> c_3(circ^#(s, t))
, circ^#(cons(lift(), s), cons(lift(), t)) -> c_4(circ^#(s, t))
, msubst^#(msubst(a, s), t) -> c_5(circ^#(s, t)) }
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))