Term Rewriting System R:
[x, y]
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

PLUS(x, s(y)) -> PLUS(x, y)
TIMES(s(x), y) -> PLUS(times(x, y), y)
TIMES(s(x), y) -> TIMES(x, y)
P(s(s(x))) -> P(s(x))
FAC(s(x)) -> TIMES(fac(p(s(x))), s(x))
FAC(s(x)) -> FAC(p(s(x)))
FAC(s(x)) -> P(s(x))

Furthermore, R contains four SCCs. 


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


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 5
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 5
             ↳FwdInst
             ...
               →DP Problem 6
                 ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

PLUS(x'''', s(s(s(y'''')))) -> PLUS(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:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 5
             ↳FwdInst
             ...
               →DP Problem 7
                 ↳Dependency Graph
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar


Dependency Pair:


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

P(s(s(x))) -> P(s(x))


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

P(s(s(x))) -> P(s(x))
one new Dependency Pair is created:

P(s(s(s(x'')))) -> P(s(s(x'')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 8
             ↳Forward Instantiation Transformation
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar


Dependency Pair:

P(s(s(s(x'')))) -> P(s(s(x'')))


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

P(s(s(s(x'')))) -> P(s(s(x'')))
one new Dependency Pair is created:

P(s(s(s(s(x''''))))) -> P(s(s(s(x''''))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 8
             ↳FwdInst
             ...
               →DP Problem 9
                 ↳Argument Filtering and Ordering
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar


Dependency Pair:

P(s(s(s(s(x''''))))) -> P(s(s(s(x''''))))


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

P(s(s(s(s(x''''))))) -> P(s(s(s(x''''))))


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:
P(x1) -> P(x1)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
           →DP Problem 8
             ↳FwdInst
             ...
               →DP Problem 10
                 ↳Dependency Graph
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar


Dependency Pair:


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
           →DP Problem 11
             ↳Forward Instantiation Transformation
       →DP Problem 4
         ↳Nar


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

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

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
           →DP Problem 11
             ↳FwdInst
             ...
               →DP Problem 12
                 ↳Argument Filtering and Ordering
       →DP Problem 4
         ↳Nar


Dependency Pair:

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


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




The following dependency pair can be strictly oriented:

TIMES(s(s(s(x''''))), y'''') -> TIMES(s(s(x'''')), 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:
TIMES(x1, x2) -> TIMES(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
           →DP Problem 11
             ↳FwdInst
             ...
               →DP Problem 13
                 ↳Dependency Graph
       →DP Problem 4
         ↳Nar


Dependency Pair:


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Narrowing Transformation


Dependency Pair:

FAC(s(x)) -> FAC(p(s(x)))


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

FAC(s(x)) -> FAC(p(s(x)))
two new Dependency Pairs are created:

FAC(s(s(x''))) -> FAC(s(p(s(x''))))
FAC(s(0)) -> FAC(0)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar
           →DP Problem 14
             ↳Narrowing Transformation


Dependency Pair:

FAC(s(s(x''))) -> FAC(s(p(s(x''))))


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

FAC(s(s(x''))) -> FAC(s(p(s(x''))))
two new Dependency Pairs are created:

FAC(s(s(s(x')))) -> FAC(s(s(p(s(x')))))
FAC(s(s(0))) -> FAC(s(0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 15
                 ↳Narrowing Transformation


Dependency Pair:

FAC(s(s(s(x')))) -> FAC(s(s(p(s(x')))))


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

FAC(s(s(s(x')))) -> FAC(s(s(p(s(x')))))
two new Dependency Pairs are created:

FAC(s(s(s(s(x''))))) -> FAC(s(s(s(p(s(x''))))))
FAC(s(s(s(0)))) -> FAC(s(s(0)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 16
                 ↳Narrowing Transformation


Dependency Pair:

FAC(s(s(s(s(x''))))) -> FAC(s(s(s(p(s(x''))))))


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

FAC(s(s(s(s(x''))))) -> FAC(s(s(s(p(s(x''))))))
two new Dependency Pairs are created:

FAC(s(s(s(s(s(x')))))) -> FAC(s(s(s(s(p(s(x')))))))
FAC(s(s(s(s(0))))) -> FAC(s(s(s(0))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 17
                 ↳Narrowing Transformation


Dependency Pair:

FAC(s(s(s(s(s(x')))))) -> FAC(s(s(s(s(p(s(x')))))))


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost




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

FAC(s(s(s(s(s(x')))))) -> FAC(s(s(s(s(p(s(x')))))))
two new Dependency Pairs are created:

FAC(s(s(s(s(s(s(x''))))))) -> FAC(s(s(s(s(s(p(s(x''))))))))
FAC(s(s(s(s(s(0)))))) -> FAC(s(s(s(s(0)))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳FwdInst
       →DP Problem 3
         ↳FwdInst
       →DP Problem 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 18
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

FAC(s(s(s(s(s(s(x''))))))) -> FAC(s(s(s(s(s(p(s(x''))))))))


Rules:


plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))
times(0, y) -> 0
times(x, 0) -> 0
times(s(x), y) -> plus(times(x, y), y)
p(s(s(x))) -> s(p(s(x)))
p(s(0)) -> 0
fac(s(x)) -> times(fac(p(s(x))), s(x))


Strategy:

innermost



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