Term Rewriting System R:
[x, y]
-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
+'(s(x), y) -> +'(x, y)
*'(x, s(y)) -> +'(x, *(x, y))
*'(x, s(y)) -> *'(x, y)
F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))
F(s(x)) -> -'(*(s(s(0)), s(x)), s(s(x)))
F(s(x)) -> *'(s(s(0)), s(x))

Furthermore, R contains four SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




As we are in the innermost case, we can delete all 7 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 5
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. -'(s(x), s(y)) -> -'(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




As we are in the innermost case, we can delete all 7 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 6
             ↳Size-Change Principle
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. +'(s(x), y) -> +'(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2=2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1>1
2=2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳Usable Rules (Innermost)
       →DP Problem 4
         ↳UsableRules


Dependency Pair:

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


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




As we are in the innermost case, we can delete all 7 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
           →DP Problem 7
             ↳Size-Change Principle
       →DP Problem 4
         ↳UsableRules


Dependency Pair:

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


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. *'(x, s(y)) -> *'(x, y)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2>2

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1=1
2>2

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳Usable Rules (Innermost)


Dependency Pair:

F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))


Rules:


-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
*(x, 0) -> 0
*(x, s(y)) -> +(x, *(x, y))
f(s(x)) -> f(-(*(s(s(0)), s(x)), s(s(x))))


Strategy:

innermost




As we are in the innermost case, we can delete all 1 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(x)) -> F(-(*(s(s(0)), s(x)), s(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 9
                 ↳Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(x)) -> F(-(+(s(s(0)), *(s(s(0)), x)), s(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 10
                 ↳Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(x)) -> F(-(s(+(s(0), *(s(s(0)), x))), s(s(x))))
one new Dependency Pair is created:

F(s(x)) -> F(-(+(s(0), *(s(s(0)), x)), s(x)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 11
                 ↳Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(+(s(0), *(s(s(0)), x)), s(x)))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(x)) -> F(-(+(s(0), *(s(s(0)), x)), s(x)))
one new Dependency Pair is created:

F(s(x)) -> F(-(s(+(0, *(s(s(0)), x))), s(x)))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 12
                 ↳Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(s(+(0, *(s(s(0)), x))), s(x)))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(x)) -> F(-(s(+(0, *(s(s(0)), x))), s(x)))
one new Dependency Pair is created:

F(s(x)) -> F(-(+(0, *(s(s(0)), x)), x))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 13
                 ↳Rewriting Transformation


Dependency Pair:

F(s(x)) -> F(-(+(0, *(s(s(0)), x)), x))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(x)) -> F(-(+(0, *(s(s(0)), x)), x))
one new Dependency Pair is created:

F(s(x)) -> F(-(*(s(s(0)), x), x))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 14
                 ↳Narrowing Transformation


Dependency Pair:

F(s(x)) -> F(-(*(s(s(0)), x), x))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(x)) -> F(-(*(s(s(0)), x), x))
three new Dependency Pairs are created:

F(s(0)) -> F(*(s(s(0)), 0))
F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), s(y')))
F(s(0)) -> F(-(0, 0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 15
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(0)) -> F(-(0, 0))
F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), s(y')))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(s(y'))) -> F(-(+(s(s(0)), *(s(s(0)), y')), s(y')))
one new Dependency Pair is created:

F(s(s(y'))) -> F(-(s(+(s(0), *(s(s(0)), y'))), s(y')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 16
                 ↳Rewriting Transformation


Dependency Pairs:

F(s(s(y'))) -> F(-(s(+(s(0), *(s(s(0)), y'))), s(y')))
F(s(0)) -> F(-(0, 0))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(0)) -> F(-(0, 0))
one new Dependency Pair is created:

F(s(0)) -> F(0)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 17
                 ↳Rewriting Transformation


Dependency Pair:

F(s(s(y'))) -> F(-(s(+(s(0), *(s(s(0)), y'))), s(y')))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(s(y'))) -> F(-(s(+(s(0), *(s(s(0)), y'))), s(y')))
one new Dependency Pair is created:

F(s(s(y'))) -> F(-(+(s(0), *(s(s(0)), y')), y'))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 18
                 ↳Rewriting Transformation


Dependency Pair:

F(s(s(y'))) -> F(-(+(s(0), *(s(s(0)), y')), y'))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(s(y'))) -> F(-(+(s(0), *(s(s(0)), y')), y'))
one new Dependency Pair is created:

F(s(s(y'))) -> F(-(s(+(0, *(s(s(0)), y'))), y'))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 19
                 ↳Rewriting Transformation


Dependency Pair:

F(s(s(y'))) -> F(-(s(+(0, *(s(s(0)), y'))), y'))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost




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

F(s(s(y'))) -> F(-(s(+(0, *(s(s(0)), y'))), y'))
one new Dependency Pair is created:

F(s(s(y'))) -> F(-(s(*(s(s(0)), y')), y'))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳UsableRules
           →DP Problem 8
             ↳Rw
             ...
               →DP Problem 20
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pair:

F(s(s(y'))) -> F(-(s(*(s(s(0)), y')), y'))


Rules:


+(0, y) -> y
+(s(x), y) -> s(+(x, y))
-(s(x), s(y)) -> -(x, y)
-(x, 0) -> x
*(x, s(y)) -> +(x, *(x, y))
*(x, 0) -> 0


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:27 minutes