We consider the following Problem:
Strict Trs:
{ f(empty(), l) -> l
, f(cons(x, k), l) -> g(k, l, cons(x, k))
, g(a, b, c) -> f(a, cons(b, c))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ f(empty(), l) -> l
, f(cons(x, k), l) -> g(k, l, cons(x, k))
, g(a, b, c) -> f(a, cons(b, c))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {f(empty(), l) -> l}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(f) = {}, Uargs(cons) = {}, Uargs(g) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
f(x1, x2) = [0 0] x1 + [1 0] x2 + [1]
[0 0] [1 1] [1]
empty() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
g(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 0] x3 + [1]
[1 1] [0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ f(cons(x, k), l) -> g(k, l, cons(x, k))
, g(a, b, c) -> f(a, cons(b, c))}
Weak Trs: {f(empty(), l) -> l}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {g(a, b, c) -> f(a, cons(b, c))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(f) = {}, Uargs(cons) = {}, Uargs(g) = {}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
f(x1, x2) = [0 0] x1 + [1 0] x2 + [0]
[0 0] [0 1] [1]
empty() = [0]
[0]
cons(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
g(x1, x2, x3) = [0 0] x1 + [1 0] x2 + [0 0] x3 + [3]
[1 0] [0 1] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs: {f(cons(x, k), l) -> g(k, l, cons(x, k))}
Weak Trs:
{ g(a, b, c) -> f(a, cons(b, c))
, f(empty(), l) -> l}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs: {f(cons(x, k), l) -> g(k, l, cons(x, k))}
Weak Trs:
{ g(a, b, c) -> f(a, cons(b, c))
, f(empty(), l) -> l}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We have computed the following dependency pairs
Strict DPs: {f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
Weak DPs:
{ g^#(a, b, c) -> f^#(a, cons(b, c))
, f^#(empty(), l) -> c_3()}
We consider the following Problem:
Strict DPs: {f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
Strict Trs: {f(cons(x, k), l) -> g(k, l, cons(x, k))}
Weak DPs:
{ g^#(a, b, c) -> f^#(a, cons(b, c))
, f^#(empty(), l) -> c_3()}
Weak Trs:
{ g(a, b, c) -> f(a, cons(b, c))
, f(empty(), l) -> l}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
No rule is usable.
We consider the following Problem:
Strict DPs: {f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
Weak DPs:
{ g^#(a, b, c) -> f^#(a, cons(b, c))
, f^#(empty(), l) -> c_3()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
Weak DPs:
{ g^#(a, b, c) -> f^#(a, cons(b, c))
, f^#(empty(), l) -> c_3()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We use following congruence DG for path analysis
->1:{1,2} [ YES(O(1),O(1)) ]
|
`->2:{3} [ YES(O(1),O(1)) ]
Here dependency-pairs are as follows:
Strict DPs:
{1: f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
WeakDPs DPs:
{ 2: g^#(a, b, c) -> f^#(a, cons(b, c))
, 3: f^#(empty(), l) -> c_3()}
* Path 1:{1,2}: YES(O(1),O(1))
----------------------------
We consider the following Problem:
Strict DPs: {f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))
together with the congruence-graph
->1:{1} Noncyclic, trivial, SCC
Here dependency-pairs are as follows:
Strict DPs:
{1: f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 1:{1,2}->2:{3}: YES(O(1),O(1))
-----------------------------------
We consider the following Problem:
Weak DPs:
{ g^#(a, b, c) -> f^#(a, cons(b, c))
, f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: g^#(a, b, c) -> f^#(a, cons(b, c))
-->_1 f^#(cons(x, k), l) -> g^#(k, l, cons(x, k)) :2
2: f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))
-->_1 g^#(a, b, c) -> f^#(a, cons(b, c)) :1
together with the congruence-graph
->1:{1,2} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{ 1: g^#(a, b, c) -> f^#(a, cons(b, c))
, 2: f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{ 1: g^#(a, b, c) -> f^#(a, cons(b, c))
, 2: f^#(cons(x, k), l) -> g^#(k, l, cons(x, k))}
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
Hurray, we answered YES(?,O(n^1))