We consider the following Problem:
Strict Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()
, le(s(x), s(y)) -> le(x, y)
, quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
, if_quot(true(), x, y) -> s(quot(minus(x, y), y))
, if_quot(false(), x, y) -> 0()}
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)) -> minus(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()
, le(s(x), s(y)) -> le(x, y)
, quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
, if_quot(true(), x, y) -> s(quot(minus(x, y), y))
, if_quot(false(), x, 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:
{ le(0(), y) -> true()
, le(s(x), 0()) -> false()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
Uargs(quot) = {1}, Uargs(if_quot) = {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]
[0 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
le(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [0]
[0]
quot(x1, x2) = [1 2] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
if_quot(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [0 0] x3 + [0]
[0 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
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
, if_quot(true(), x, y) -> s(quot(minus(x, y), y))
, if_quot(false(), x, y) -> 0()}
Weak Trs:
{ le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {if_quot(false(), x, y) -> 0()}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
Uargs(quot) = {1}, Uargs(if_quot) = {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]
[0 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
le(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
true() = [0]
[0]
false() = [1]
[0]
quot(x1, x2) = [1 2] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
if_quot(x1, x2, x3) = [1 0] x1 + [1 1] x2 + [0 0] x3 + [0]
[0 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
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))
, if_quot(true(), x, y) -> s(quot(minus(x, y), y))}
Weak Trs:
{ if_quot(false(), x, y) -> 0()
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {if_quot(true(), x, y) -> s(quot(minus(x, y), y))}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
Uargs(quot) = {1}, Uargs(if_quot) = {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]
[0 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 0] [1]
le(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
true() = [1]
[1]
false() = [0]
[0]
quot(x1, x2) = [1 2] x1 + [0 0] x2 + [1]
[0 0] [0 0] [1]
if_quot(x1, x2, x3) = [1 3] x1 + [1 1] x2 + [0 1] x3 + [2]
[0 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
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))}
Weak Trs:
{ if_quot(true(), x, y) -> s(quot(minus(x, y), y))
, if_quot(false(), x, y) -> 0()
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
The weightgap principle applies, where following rules are oriented strictly:
TRS Component: {le(s(x), s(y)) -> le(x, y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
Uargs(quot) = {1}, Uargs(if_quot) = {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 0] [0 0] [0]
0() = [0]
[0]
s(x1) = [1 0] x1 + [0]
[0 1] [2]
le(x1, x2) = [0 0] x1 + [0 1] x2 + [0]
[0 0] [0 0] [3]
true() = [0]
[2]
false() = [0]
[2]
quot(x1, x2) = [1 1] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
if_quot(x1, x2, x3) = [1 2] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 0] [0 1] [0 0] [3]
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)) -> minus(x, y)
, quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))}
Weak Trs:
{ le(s(x), s(y)) -> le(x, y)
, if_quot(true(), x, y) -> s(quot(minus(x, y), y))
, if_quot(false(), x, y) -> 0()
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
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)) -> minus(x, y)}
Interpretation of nonconstant growth:
-------------------------------------
The following argument positions are usable:
Uargs(minus) = {}, Uargs(s) = {1}, Uargs(le) = {},
Uargs(quot) = {1}, Uargs(if_quot) = {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]
[0 0] [0 0] [1]
0() = [0]
[0]
s(x1) = [1 0] x1 + [1]
[0 0] [0]
le(x1, x2) = [0 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [3]
true() = [0]
[3]
false() = [0]
[0]
quot(x1, x2) = [1 1] x1 + [0 0] x2 + [0]
[0 0] [0 0] [1]
if_quot(x1, x2, x3) = [1 1] x1 + [1 1] x2 + [0 0] x3 + [0]
[0 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
, quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))}
Weak Trs:
{ minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, if_quot(true(), x, y) -> s(quot(minus(x, y), y))
, if_quot(false(), x, y) -> 0()
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
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(le) = {},
Uargs(quot) = {1}, Uargs(if_quot) = {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]
[0 1] [0 0] [0]
0() = [0]
[1]
s(x1) = [1 0] x1 + [0]
[0 1] [0]
le(x1, x2) = [0 0] x1 + [0 0] x2 + [1]
[0 1] [0 0] [3]
true() = [1]
[0]
false() = [0]
[0]
quot(x1, x2) = [1 0] x1 + [0 0] x2 + [0]
[0 0] [0 0] [0]
if_quot(x1, x2, x3) = [1 0] x1 + [1 0] x2 + [0 0] x3 + [0]
[0 0] [0 0] [0 0] [1]
The strictly oriented rules are moved into the weak component.
We consider the following Problem:
Strict Trs: {quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, if_quot(true(), x, y) -> s(quot(minus(x, y), y))
, if_quot(false(), x, y) -> 0()
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict Trs: {quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, if_quot(true(), x, y) -> s(quot(minus(x, y), y))
, if_quot(false(), x, y) -> 0()
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We have computed the following dependency pairs
Strict DPs: {quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
Weak DPs:
{ minus^#(x, 0()) -> c_2()
, minus^#(s(x), s(y)) -> minus^#(x, y)
, le^#(s(x), s(y)) -> le^#(x, y)
, if_quot^#(true(), x, y) -> quot^#(minus(x, y), y)
, if_quot^#(false(), x, y) -> c_6()
, le^#(0(), y) -> c_7()
, le^#(s(x), 0()) -> c_8()}
We consider the following Problem:
Strict DPs: {quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
Strict Trs: {quot(x, s(y)) -> if_quot(le(s(y), x), x, s(y))}
Weak DPs:
{ minus^#(x, 0()) -> c_2()
, minus^#(s(x), s(y)) -> minus^#(x, y)
, le^#(s(x), s(y)) -> le^#(x, y)
, if_quot^#(true(), x, y) -> quot^#(minus(x, y), y)
, if_quot^#(false(), x, y) -> c_6()
, le^#(0(), y) -> c_7()
, le^#(s(x), 0()) -> c_8()}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, if_quot(true(), x, y) -> s(quot(minus(x, y), y))
, if_quot(false(), x, y) -> 0()
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We replace strict/weak-rules by the corresponding usable rules:
Weak Usable Rules:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
We consider the following Problem:
Strict DPs: {quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
Weak DPs:
{ minus^#(x, 0()) -> c_2()
, minus^#(s(x), s(y)) -> minus^#(x, y)
, le^#(s(x), s(y)) -> le^#(x, y)
, if_quot^#(true(), x, y) -> quot^#(minus(x, y), y)
, if_quot^#(false(), x, y) -> c_6()
, le^#(0(), y) -> c_7()
, le^#(s(x), 0()) -> c_8()}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We consider the following Problem:
Strict DPs: {quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
Weak DPs:
{ minus^#(x, 0()) -> c_2()
, minus^#(s(x), s(y)) -> minus^#(x, y)
, le^#(s(x), s(y)) -> le^#(x, y)
, if_quot^#(true(), x, y) -> quot^#(minus(x, y), y)
, if_quot^#(false(), x, y) -> c_6()
, le^#(0(), y) -> c_7()
, le^#(s(x), 0()) -> c_8()}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(?,O(n^1))
Proof:
We use following congruence DG for path analysis
->6:{1,5} [ YES(O(1),O(1)) ]
|
`->7:{6} [ YES(O(1),O(1)) ]
->4:{3} [ subsumed ]
|
`->5:{2} [ YES(O(1),O(1)) ]
->1:{4} [ subsumed ]
|
|->2:{7} [ YES(O(1),O(1)) ]
|
`->3:{8} [ YES(O(1),O(1)) ]
Here dependency-pairs are as follows:
Strict DPs:
{1: quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
WeakDPs DPs:
{ 2: minus^#(x, 0()) -> c_2()
, 3: minus^#(s(x), s(y)) -> minus^#(x, y)
, 4: le^#(s(x), s(y)) -> le^#(x, y)
, 5: if_quot^#(true(), x, y) -> quot^#(minus(x, y), y)
, 6: if_quot^#(false(), x, y) -> c_6()
, 7: le^#(0(), y) -> c_7()
, 8: le^#(s(x), 0()) -> c_8()}
* Path 6:{1,5}: YES(O(1),O(1))
----------------------------
We consider the following Problem:
Strict DPs: {quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))
together with the congruence-graph
->1:{1} Noncyclic, trivial, SCC
Here dependency-pairs are as follows:
Strict DPs:
{1: quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
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 6:{1,5}->7:{6}: YES(O(1),O(1))
-----------------------------------
We consider the following Problem:
Weak DPs:
{ if_quot^#(true(), x, y) -> quot^#(minus(x, y), y)
, quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: if_quot^#(true(), x, y) -> quot^#(minus(x, y), y)
-->_1 quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y)) :2
2: quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))
-->_1 if_quot^#(true(), x, y) -> quot^#(minus(x, y), y) :1
together with the congruence-graph
->1:{1,2} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{ 1: if_quot^#(true(), x, y) -> quot^#(minus(x, y), y)
, 2: quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{ 1: if_quot^#(true(), x, y) -> quot^#(minus(x, y), y)
, 2: quot^#(x, s(y)) -> if_quot^#(le(s(y), x), x, s(y))}
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
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 4:{3}: subsumed
--------------------
This path is subsumed by the proof of paths 4:{3}->5:{2}.
* Path 4:{3}->5:{2}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Weak DPs: {minus^#(s(x), s(y)) -> minus^#(x, y)}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: minus^#(s(x), s(y)) -> minus^#(x, y)
-->_1 minus^#(s(x), s(y)) -> minus^#(x, y) :1
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: minus^#(s(x), s(y)) -> 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)) -> minus^#(x, y)}
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
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:{8},
1:{4}->2:{7}.
* Path 1:{4}->2:{7}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Weak DPs: {le^#(s(x), s(y)) -> le^#(x, y)}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: le^#(s(x), s(y)) -> le^#(x, y)
-->_1 le^#(s(x), s(y)) -> le^#(x, y) :1
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: le^#(s(x), s(y)) -> le^#(x, y)}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: le^#(s(x), s(y)) -> le^#(x, y)}
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
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}->3:{8}: YES(O(1),O(1))
---------------------------------
We consider the following Problem:
Weak DPs: {le^#(s(x), s(y)) -> le^#(x, y)}
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the the dependency-graph
1: le^#(s(x), s(y)) -> le^#(x, y)
-->_1 le^#(s(x), s(y)) -> le^#(x, y) :1
together with the congruence-graph
->1:{1} Weak SCC
Here dependency-pairs are as follows:
WeakDPs DPs:
{1: le^#(s(x), s(y)) -> le^#(x, y)}
The following rules are either leafs or part of trailing weak paths, and thus they can be removed:
{1: le^#(s(x), s(y)) -> le^#(x, y)}
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
StartTerms: basic terms
Strategy: innermost
Certificate: YES(O(1),O(1))
Proof:
We consider the following Problem:
Weak Trs:
{ minus(x, 0()) -> x
, minus(s(x), s(y)) -> minus(x, y)
, le(s(x), s(y)) -> le(x, y)
, le(0(), y) -> true()
, le(s(x), 0()) -> false()}
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))