Term Rewriting System R:
[x, y]
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

MINUS(x, s(y)) -> PRED(minus(x, y))
MINUS(x, s(y)) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(x, y)
LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))
LOG(s(s(x))) -> QUOT(x, s(s(0)))

Furthermore, R contains three SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pair:

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


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




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

MINUS(x, s(y)) -> MINUS(x, y)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 4
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pair:

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


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




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

MINUS(x'', s(s(y''))) -> MINUS(x'', s(y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 4
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pair:

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


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


There are no usable rules for innermost w.r.t. to the AFS 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
         ↳FwdInst
           →DP Problem 4
             ↳FwdInst
             ...
               →DP Problem 6
                 ↳Dependency Graph
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pair:


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Narrowing Transformation
       →DP Problem 3
         ↳Nar


Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




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

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

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

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 7
             ↳Forward Instantiation Transformation
           →DP Problem 8
             ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pair:

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




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

QUOT(s(x''), s(0)) -> QUOT(x'', s(0))
one new Dependency Pair is created:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 7
             ↳FwdInst
             ...
               →DP Problem 9
                 ↳Argument Filtering and Ordering
           →DP Problem 8
             ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pair:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(s(x'''')), s(0)) -> QUOT(s(x''''), s(0))


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


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 7
             ↳FwdInst
             ...
               →DP Problem 12
                 ↳Dependency Graph
           →DP Problem 8
             ↳Nar
       →DP Problem 3
         ↳Nar


Dependency Pair:


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 7
             ↳FwdInst
           →DP Problem 8
             ↳Narrowing Transformation
       →DP Problem 3
         ↳Nar


Dependency Pair:

QUOT(s(x''), s(s(y''))) -> QUOT(pred(minus(x'', y'')), s(s(y'')))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




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

QUOT(s(x''), s(s(y''))) -> QUOT(pred(minus(x'', y'')), s(s(y'')))
two new Dependency Pairs are created:

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

The transformation is resulting in two new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 7
             ↳FwdInst
           →DP Problem 8
             ↳Nar
             ...
               →DP Problem 10
                 ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳Nar


Dependency Pair:

QUOT(s(x'''), s(s(0))) -> QUOT(pred(x'''), s(s(0)))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(x'''), s(s(0))) -> QUOT(pred(x'''), s(s(0)))


The following usable rule for innermost w.r.t. to the AFS can be oriented:

pred(s(x)) -> x


Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
pred(x1) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 7
             ↳FwdInst
           →DP Problem 8
             ↳Nar
             ...
               →DP Problem 11
                 ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳Nar


Dependency Pair:

QUOT(s(x'''), s(s(s(y')))) -> QUOT(pred(pred(minus(x''', y'))), s(s(s(y'))))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

QUOT(s(x'''), s(s(s(y')))) -> QUOT(pred(pred(minus(x''', y'))), s(s(s(y'))))


The following usable rules for innermost w.r.t. to the AFS can be oriented:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))


Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
QUOT(x1, x2) -> QUOT(x1, x2)
s(x1) -> s(x1)
pred(x1) -> x1
minus(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Narrowing Transformation


Dependency Pair:

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




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

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))
two new Dependency Pairs are created:

LOG(s(s(0))) -> LOG(s(0))
LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar
           →DP Problem 15
             ↳Rewriting Transformation


Dependency Pair:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




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

LOG(s(s(s(x'')))) -> LOG(s(s(quot(minus(x'', s(0)), s(s(0))))))
one new Dependency Pair is created:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(minus(x'', 0)), s(s(0))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar
           →DP Problem 15
             ↳Rw
             ...
               →DP Problem 16
                 ↳Rewriting Transformation


Dependency Pair:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(minus(x'', 0)), s(s(0))))))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




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

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(minus(x'', 0)), s(s(0))))))
one new Dependency Pair is created:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(x''), s(s(0))))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar
           →DP Problem 15
             ↳Rw
             ...
               →DP Problem 17
                 ↳Argument Filtering and Ordering


Dependency Pair:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(x''), s(s(0))))))


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




The following dependency pair can be strictly oriented:

LOG(s(s(s(x'')))) -> LOG(s(s(quot(pred(x''), s(s(0))))))


The following usable rules for innermost w.r.t. to the AFS can be oriented:

pred(s(x)) -> x
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))


Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:
LOG(x1) -> LOG(x1)
s(x1) -> s(x1)
quot(x1, x2) -> x1
pred(x1) -> x1
minus(x1, x2) -> x1


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
       →DP Problem 3
         ↳Nar
           →DP Problem 15
             ↳Rw
             ...
               →DP Problem 18
                 ↳Dependency Graph


Dependency Pair:


Rules:


pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes