Term Rewriting System R:
[x, y]
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)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0

Innermost Termination of R to be shown.



   R
Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
QUOT(x, s(y)) -> IFQUOT(le(s(y), x), x, s(y))
QUOT(x, s(y)) -> LE(s(y), x)
IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
IFQUOT(true, x, y) -> MINUS(x, y)

Furthermore, R contains three SCCs.


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


Dependency Pair:

MINUS(s(x), s(y)) -> MINUS(x, y)


Rules:


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)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




The following dependency pair can be strictly oriented:

MINUS(s(x), s(y)) -> MINUS(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:
MINUS(x1, x2) -> MINUS(x1, x2)
s(x1) -> s(x1)


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


Dependency Pair:


Rules:


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)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)


Rules:


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)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


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 2
AFS
           →DP Problem 5
Dependency Graph
       →DP Problem 3
Nar


Dependency Pair:


Rules:


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)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IFQUOT(le(s(y), x), x, s(y))


Rules:


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)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




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

QUOT(x, s(y)) -> IFQUOT(le(s(y), x), x, s(y))
two new Dependency Pairs are created:

QUOT(0, s(y')) -> IFQUOT(false, 0, s(y'))
QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 6
Narrowing Transformation


Dependency Pairs:

QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))
IFQUOT(true, x, y) -> QUOT(minus(x, y), y)


Rules:


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)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost




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

IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
two new Dependency Pairs are created:

IFQUOT(true, x'', 0) -> QUOT(x'', 0)
IFQUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))

The transformation is resulting in one new DP problem:



   R
DPs
       →DP Problem 1
AFS
       →DP Problem 2
AFS
       →DP Problem 3
Nar
           →DP Problem 6
Nar
             ...
               →DP Problem 7
Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

IFQUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))
QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))


Rules:


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)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0


Strategy:

innermost



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