We consider the following Problem:
Strict Trs:
{ minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))
, p(s(X)) -> X
, div(0(), s(Y)) -> 0()
, div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs:
{ minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))
, p(s(X)) -> X
, div(0(), s(Y)) -> 0()
, div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
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(), s(Y)) -> 0()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {1}, Uargs(p) = {1}, Uargs(div) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
minus(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[1 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
p(x1) = [1 0] x1 + [0]
[0 0] [1]
div(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))
, p(s(X)) -> X
, div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
Weak Trs: {div(0(), s(Y)) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {minus(s(X), s(Y)) -> p(minus(X, Y))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {1}, Uargs(p) = {1}, Uargs(div) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
minus(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[1 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [2]
[0 0] [1]
p(x1) = [1 0] x1 + [0]
[0 0] [1]
div(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ minus(X, 0()) -> X
, p(s(X)) -> X
, div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
Weak Trs:
{ minus(s(X), s(Y)) -> p(minus(X, Y))
, div(0(), s(Y)) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {minus(X, 0()) -> X}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {1}, Uargs(p) = {1}, Uargs(div) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
minus(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[1 1] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
p(x1) = [1 0] x1 + [0]
[1 0] [1]
div(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs:
{ p(s(X)) -> X
, div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
Weak Trs:
{ minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))
, div(0(), s(Y)) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {p(s(X)) -> X}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {1}, Uargs(p) = {1}, Uargs(div) = {1}
We have the following EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
minus(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 1] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [1]
[0 1] [0]
p(x1) = [1 0] x1 + [1]
[0 1] [0]
div(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [1 0] [0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs: {div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))
, div(0(), s(Y)) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs: {div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))
, div(0(), s(Y)) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We have computed the following dependency pairs
Strict DPs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
Weak DPs:
{ p^#(s(X)) -> c_2()
, minus^#(X, 0()) -> c_3()
, minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
, div^#(0(), s(Y)) -> c_5()}
We consider the following Problem:
Strict DPs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
Strict Trs: {div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))}
Weak DPs:
{ p^#(s(X)) -> c_2()
, minus^#(X, 0()) -> c_3()
, minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
, div^#(0(), s(Y)) -> c_5()}
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))
, div(0(), s(Y)) -> 0()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We replace strict/weak-rules by the corresponding usable rules:
Weak Usable Rules:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
We consider the following Problem:
Strict DPs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
Weak DPs:
{ p^#(s(X)) -> c_2()
, minus^#(X, 0()) -> c_3()
, minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
, div^#(0(), s(Y)) -> c_5()}
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
Dependency Pairs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
Interpretation of constant growth:
----------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {}, Uargs(p) = {1}, Uargs(div) = {},
Uargs(div^#) = {1}, Uargs(p^#) = {1}, Uargs(minus^#) = {}
We have the following constructor-based EDA-non-satisfying and IDA(1)-non-satisfying matrix interpretation:
Interpretation Functions:
minus(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 1] [1 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [2]
[0 1] [0]
p(x1) = [1 0] x1 + [1]
[0 1] [1]
div(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
div^#(x1, x2) = [1 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
p^#(x1) = [1 0] x1 + [1]
[0 0] [1]
c_2() = [0]
[0]
minus^#(x1, x2) = [1 0] x1 + [1 0] x2 + [1]
[0 0] [0 0] [1]
c_3() = [0]
[0]
c_5() = [0]
[0]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Weak DPs:
{ div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))
, p^#(s(X)) -> c_2()
, minus^#(X, 0()) -> c_3()
, minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
, div^#(0(), s(Y)) -> c_5()}
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We use following congruence DG for path analysis
->4:{1} [ subsumed ]
|
`->5:{5} [ YES(O(1),O(1)) ]
->2:{3} [ YES(O(1),O(1)) ]
->1:{4} [ subsumed ]
|
`->3:{2} [ YES(O(1),O(1)) ]
Here dependency-pairs are as follows:
WeakDPs DPs:
{ 1: div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))
, 2: p^#(s(X)) -> c_2()
, 3: minus^#(X, 0()) -> c_3()
, 4: minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
, 5: div^#(0(), s(Y)) -> c_5()}
* Path 4:{1}: subsumed
--------------------
This path is subsumed by the proof of paths 4:{1}->5:{5}.
* Path 4:{1}->5:{5}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Weak DPs: {div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))
-->_1 div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y)) :1
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: div^#(s(X), s(Y)) -> div^#(minus(X, Y), s(Y))}
We consider the following Problem:
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 2:{3}: YES(O(1),O(1))
--------------------------
We consider the following Problem:
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
We consider the following Problem:
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
Empty rules are trivially bounded
* Path 1:{4}: subsumed
--------------------
This path is subsumed by the proof of paths 1:{4}->3:{2}.
* Path 1:{4}->3:{2}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Weak DPs: {minus^#(s(X), s(Y)) -> p^#(minus(X, Y))}
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: minus^#(s(X), s(Y)) -> p^#(minus(X, Y))
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: minus^#(s(X), s(Y)) -> p^#(minus(X, Y))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: minus^#(s(X), s(Y)) -> p^#(minus(X, Y))}
We consider the following Problem:
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ p(s(X)) -> X
, minus(X, 0()) -> X
, minus(s(X), s(Y)) -> p(minus(X, Y))}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
No rule is usable.
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))