Term Rewriting System R:
[Y, X]
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

LE(s(X), s(Y)) -> LE(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
MINUS(s(X), Y) -> LE(s(X), Y)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains three SCCs.


   R
DPs
       →DP Problem 1
Argument Filtering and Ordering
       →DP Problem 2
Nar
       →DP Problem 3
Remaining


Dependency Pair:

LE(s(X), s(Y)) -> LE(X, Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

LE(s(X), s(Y)) -> LE(X, Y)


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
LE(x1, x2) -> LE(x1, x2)
s(x1) -> s(x1)


   R
DPs
       →DP Problem 1
AFS
           →DP Problem 4
Dependency Graph
       →DP Problem 2
Nar
       →DP Problem 3
Remaining


Dependency Pair:


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Narrowing Transformation
       →DP Problem 3
Remaining


Dependency Pairs:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
two new Dependency Pairs are created:

MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)
MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 5
Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pairs:

MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
two new Dependency Pairs are created:

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
IFMINUS(false, s(X'), s(Y'''')) -> MINUS(X', s(Y''''))

The transformation is resulting in two new DP problems:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 5
Inst
             ...
               →DP Problem 6
Forward Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pairs:

IFMINUS(false, s(X'), s(Y'''')) -> MINUS(X', s(Y''''))
MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IFMINUS(false, s(X'), s(Y'''')) -> MINUS(X', s(Y''''))
one new Dependency Pair is created:

IFMINUS(false, s(s(X'''')), s(Y''''')) -> MINUS(s(X''''), s(Y'''''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 5
Inst
             ...
               →DP Problem 8
Forward Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pairs:

IFMINUS(false, s(s(X'''')), s(Y''''')) -> MINUS(s(X''''), s(Y'''''))
MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

MINUS(s(X''), s(Y'')) -> IFMINUS(le(X'', Y''), s(X''), s(Y''))
one new Dependency Pair is created:

MINUS(s(s(X'''''')), s(Y''')) -> IFMINUS(le(s(X''''''), Y'''), s(s(X'''''')), s(Y'''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
       →DP Problem 3
Remaining Obligation(s)




The following remains to be proven:


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 5
Inst
             ...
               →DP Problem 7
Forward Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pairs:

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

IFMINUS(false, s(X'), 0) -> MINUS(X', 0)
one new Dependency Pair is created:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 5
Inst
             ...
               →DP Problem 9
Forward Instantiation Transformation
       →DP Problem 3
Remaining


Dependency Pairs:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)
MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




On this DP problem, a Forward Instantiation SCC transformation can be performed.
As a result of transforming the rule

MINUS(s(X''), 0) -> IFMINUS(false, s(X''), 0)
one new Dependency Pair is created:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 5
Inst
             ...
               →DP Problem 11
Argument Filtering and Ordering
       →DP Problem 3
Remaining


Dependency Pairs:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)
IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IFMINUS(false, s(s(X'''')), 0) -> MINUS(s(X''''), 0)


There are no usable rules for innermost that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
IFMINUS(x1, x2, x3) -> x2
s(x1) -> s(x1)
MINUS(x1, x2) -> x1


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
           →DP Problem 5
Inst
             ...
               →DP Problem 12
Dependency Graph
       →DP Problem 3
Remaining


Dependency Pair:

MINUS(s(s(X'''''')), 0) -> IFMINUS(false, s(s(X'''''')), 0)


Rules:


le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
Nar
       →DP Problem 3
Remaining Obligation(s)




The following remains to be proven:

Innermost Termination of R could not be shown.
Duration:
0:00 minutes