We consider the following Problem:
Strict Trs:
{ div(0(), y) -> 0()
, div(x, y) -> quot(x, y, y)
, quot(0(), s(y), z) -> 0()
, quot(s(x), s(y), z) -> quot(x, y, z)
, quot(x, 0(), s(z)) -> s(div(x, s(z)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ div(0(), y) -> 0()
, div(x, y) -> quot(x, y, y)
, quot(0(), s(y), z) -> 0()
, quot(s(x), s(y), z) -> quot(x, y, z)
, quot(x, 0(), s(z)) -> s(div(x, s(z)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component:
{ div(0(), y) -> 0()
, quot(0(), s(y), z) -> 0()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(div) = {}, Uargs(quot) = {}, Uargs(s) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
div(x1, x2) = [0 0] x1 + [0 2] x2 + [1]
[0 0] [0 0] [1]
0() = [0]
[0]
quot(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 1] x3 + [1]
[0 0] [0 0] [0 1] [1]
s(x1) = [1 2] x1 + [0]
[0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ div(x, y) -> quot(x, y, y)
, quot(s(x), s(y), z) -> quot(x, y, z)
, quot(x, 0(), s(z)) -> s(div(x, s(z)))}
Weak Trs:
{ div(0(), y) -> 0()
, quot(0(), s(y), z) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {quot(x, 0(), s(z)) -> s(div(x, s(z)))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(div) = {}, Uargs(quot) = {}, Uargs(s) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
div(x1, x2) = [0 0] x1 + [0 2] x2 + [0]
[0 0] [0 0] [0]
0() = [0]
[0]
quot(x1, x2, x3) = [0 0] x1 + [0 0] x2 + [0 1] x3 + [1]
[0 0] [0 0] [0 1] [1]
s(x1) = [1 2] x1 + [0]
[0 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ div(x, y) -> quot(x, y, y)
, quot(s(x), s(y), z) -> quot(x, y, z)}
Weak Trs:
{ quot(x, 0(), s(z)) -> s(div(x, s(z)))
, div(0(), y) -> 0()
, quot(0(), s(y), z) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {quot(s(x), s(y), z) -> quot(x, y, z)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(div) = {}, Uargs(quot) = {}, Uargs(s) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
div(x1, x2) = [1 1] x1 + [0 2] x2 + [2]
[0 0] [0 0] [0]
0() = [0]
[0]
quot(x1, x2, x3) = [1 1] x1 + [0 0] x2 + [0 1] x3 + [3]
[0 0] [0 0] [0 0] [2]
s(x1) = [1 1] x1 + [0]
[0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs: {div(x, y) -> quot(x, y, y)}
Weak Trs:
{ quot(s(x), s(y), z) -> quot(x, y, z)
, quot(x, 0(), s(z)) -> s(div(x, s(z)))
, div(0(), y) -> 0()
, quot(0(), s(y), z) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs: {div(x, y) -> quot(x, y, y)}
Weak Trs:
{ quot(s(x), s(y), z) -> quot(x, y, z)
, quot(x, 0(), s(z)) -> s(div(x, s(z)))
, div(0(), y) -> 0()
, quot(0(), s(y), z) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We have computed the following dependency pairs
Strict DPs: {div^#(x, y) -> quot^#(x, y, y)}
Weak DPs:
{ quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, div^#(0(), y) -> c_4()
, quot^#(0(), s(y), z) -> c_5()}
We consider the following Problem:
Strict DPs: {div^#(x, y) -> quot^#(x, y, y)}
Strict Trs: {div(x, y) -> quot(x, y, y)}
Weak DPs:
{ quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, div^#(0(), y) -> c_4()
, quot^#(0(), s(y), z) -> c_5()}
Weak Trs:
{ quot(s(x), s(y), z) -> quot(x, y, z)
, quot(x, 0(), s(z)) -> s(div(x, s(z)))
, div(0(), y) -> 0()
, quot(0(), s(y), z) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
No rule is usable.
We consider the following Problem:
Strict DPs: {div^#(x, y) -> quot^#(x, y, y)}
Weak DPs:
{ quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, div^#(0(), y) -> c_4()
, quot^#(0(), s(y), z) -> c_5()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {div^#(x, y) -> quot^#(x, y, y)}
Weak DPs:
{ quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, div^#(0(), y) -> c_4()
, quot^#(0(), s(y), z) -> c_5()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We use following congruence DG for path analysis
->1:{1,3,2} [ YES(O(1),O(1)) ]
|
|->3:{4} [ YES(O(1),O(1)) ]
|
`->2:{5} [ YES(O(1),O(1)) ]
Here dependency-pairs are as follows:
Strict DPs:
{1: div^#(x, y) -> quot^#(x, y, y)}
WeakDPs DPs:
{ 2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, 3: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, 4: div^#(0(), y) -> c_4()
, 5: quot^#(0(), s(y), z) -> c_5()}
* Path 1:{1,3,2}: YES(O(1),O(1))
------------------------------
We consider the following Problem:
Strict DPs: {div^#(x, y) -> quot^#(x, y, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: div^#(x, y) -> quot^#(x, y, y)
together with the congruence-graph
->1:{1} Noncyclic, trivial, SCC
Here dependency-pairs are as follows:
Strict DPs:
{1: div^#(x, y) -> quot^#(x, y, y)}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: div^#(x, y) -> quot^#(x, y, y)}
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:{1,3,2}->3:{4}: YES(O(1),O(1))
-------------------------------------
We consider the following Problem:
Weak DPs:
{ quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, div^#(x, y) -> quot^#(x, y, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
-->_1 div^#(x, y) -> quot^#(x, y, y) :3
2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
-->_1 quot^#(s(x), s(y), z) -> quot^#(x, y, z) :2
-->_1 quot^#(x, 0(), s(z)) -> div^#(x, s(z)) :1
3: div^#(x, y) -> quot^#(x, y, y)
-->_1 quot^#(s(x), s(y), z) -> quot^#(x, y, z) :2
together with the congruence-graph
->1:{1,2,3} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{ 1: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, 2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, 3: div^#(x, y) -> quot^#(x, y, y)}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{ 1: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, 2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, 3: div^#(x, y) -> quot^#(x, y, y)}
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:{1,3,2}->2:{5}: YES(O(1),O(1))
-------------------------------------
We consider the following Problem:
Weak DPs:
{ quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, div^#(x, y) -> quot^#(x, y, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
-->_1 div^#(x, y) -> quot^#(x, y, y) :3
2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
-->_1 quot^#(s(x), s(y), z) -> quot^#(x, y, z) :2
-->_1 quot^#(x, 0(), s(z)) -> div^#(x, s(z)) :1
3: div^#(x, y) -> quot^#(x, y, y)
-->_1 quot^#(s(x), s(y), z) -> quot^#(x, y, z) :2
together with the congruence-graph
->1:{1,2,3} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{ 1: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, 2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, 3: div^#(x, y) -> quot^#(x, y, y)}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{ 1: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, 2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, 3: div^#(x, y) -> quot^#(x, y, y)}
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:{1,3,2}->2:{5}: YES(O(1),O(1))
-------------------------------------
We consider the following Problem:
Weak DPs:
{ quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, div^#(x, y) -> quot^#(x, y, y)}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
-->_1 div^#(x, y) -> quot^#(x, y, y) :3
2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
-->_1 quot^#(s(x), s(y), z) -> quot^#(x, y, z) :2
-->_1 quot^#(x, 0(), s(z)) -> div^#(x, s(z)) :1
3: div^#(x, y) -> quot^#(x, y, y)
-->_1 quot^#(s(x), s(y), z) -> quot^#(x, y, z) :2
together with the congruence-graph
->1:{1,2,3} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{ 1: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, 2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, 3: div^#(x, y) -> quot^#(x, y, y)}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{ 1: quot^#(x, 0(), s(z)) -> div^#(x, s(z))
, 2: quot^#(s(x), s(y), z) -> quot^#(x, y, z)
, 3: div^#(x, y) -> quot^#(x, y, y)}
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))