Term Rewriting System R:
[x, y]
even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

EVEN(s(s(x))) -> EVEN(x)
HALF(s(s(x))) -> HALF(x)
PLUS(s(x), y) -> PLUS(x, y)
TIMES(s(x), y) -> IFTIMES(even(s(x)), s(x), y)
TIMES(s(x), y) -> EVEN(s(x))
IFTIMES(true, s(x), y) -> PLUS(times(half(s(x)), y), times(half(s(x)), y))
IFTIMES(true, s(x), y) -> TIMES(half(s(x)), y)
IFTIMES(true, s(x), y) -> HALF(s(x))
IFTIMES(false, s(x), y) -> PLUS(y, times(x, y))
IFTIMES(false, s(x), y) -> TIMES(x, y)

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:

EVEN(s(s(x))) -> EVEN(x)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

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

EVEN(s(s(s(s(x''))))) -> EVEN(s(s(x'')))

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:

EVEN(s(s(s(s(x''))))) -> EVEN(s(s(x'')))


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

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

EVEN(s(s(s(s(s(s(x''''))))))) -> EVEN(s(s(s(s(x'''')))))

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:

EVEN(s(s(s(s(s(s(x''''))))))) -> EVEN(s(s(s(s(x'''')))))


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

EVEN(s(s(s(s(s(s(x''''))))))) -> EVEN(s(s(s(s(x'''')))))


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:
EVEN(x1) -> EVEN(x1)
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:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


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:

HALF(s(s(x))) -> HALF(x)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

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

HALF(s(s(s(s(x''))))) -> HALF(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:

HALF(s(s(s(s(x''))))) -> HALF(s(s(x'')))


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

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

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(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:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

HALF(s(s(s(s(s(s(x''''))))))) -> HALF(s(s(s(s(x'''')))))


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:
HALF(x1) -> HALF(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:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


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:

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


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

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

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

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


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

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

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

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


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

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


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


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 Pairs:

IFTIMES(false, s(x), y) -> TIMES(x, y)
IFTIMES(true, s(x), y) -> TIMES(half(s(x)), y)
TIMES(s(x), y) -> IFTIMES(even(s(x)), s(x), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

TIMES(s(x), y) -> IFTIMES(even(s(x)), s(x), y)
two new Dependency Pairs are created:

TIMES(s(0), y) -> IFTIMES(false, s(0), y)
TIMES(s(s(x'')), y) -> IFTIMES(even(x''), 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 4
         ↳Nar
           →DP Problem 14
             ↳Narrowing Transformation


Dependency Pairs:

IFTIMES(true, s(x), y) -> TIMES(half(s(x)), y)
TIMES(s(s(x'')), y) -> IFTIMES(even(x''), s(s(x'')), y)
TIMES(s(0), y) -> IFTIMES(false, s(0), y)
IFTIMES(false, s(x), y) -> TIMES(x, y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

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

IFTIMES(true, s(s(x'')), y) -> TIMES(s(half(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 15
                 ↳Narrowing Transformation


Dependency Pairs:

IFTIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)
TIMES(s(0), y) -> IFTIMES(false, s(0), y)
IFTIMES(false, s(x), y) -> TIMES(x, y)
TIMES(s(s(x'')), y) -> IFTIMES(even(x''), s(s(x'')), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

TIMES(s(s(x'')), y) -> IFTIMES(even(x''), s(s(x'')), y)
three new Dependency Pairs are created:

TIMES(s(s(0)), y) -> IFTIMES(true, s(s(0)), y)
TIMES(s(s(s(0))), y) -> IFTIMES(false, s(s(s(0))), y)
TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 16
                 ↳Narrowing Transformation


Dependency Pairs:

TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)
TIMES(s(s(s(0))), y) -> IFTIMES(false, s(s(s(0))), y)
TIMES(s(s(0)), y) -> IFTIMES(true, s(s(0)), y)
IFTIMES(false, s(x), y) -> TIMES(x, y)
TIMES(s(0), y) -> IFTIMES(false, s(0), y)
IFTIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

IFTIMES(true, s(s(x'')), y) -> TIMES(s(half(x'')), y)
two new Dependency Pairs are created:

IFTIMES(true, s(s(0)), y) -> TIMES(s(0), y)
IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 17
                 ↳Instantiation Transformation


Dependency Pairs:

IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)
TIMES(s(s(s(0))), y) -> IFTIMES(false, s(s(s(0))), y)
IFTIMES(true, s(s(0)), y) -> TIMES(s(0), y)
TIMES(s(s(0)), y) -> IFTIMES(true, s(s(0)), y)
TIMES(s(0), y) -> IFTIMES(false, s(0), y)
IFTIMES(false, s(x), y) -> TIMES(x, y)
TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

IFTIMES(false, s(x), y) -> TIMES(x, y)
three new Dependency Pairs are created:

IFTIMES(false, s(0), y'') -> TIMES(0, y'')
IFTIMES(false, s(s(s(0))), y'') -> TIMES(s(s(0)), y'')
IFTIMES(false, s(s(s(s(x'0')))), y'') -> TIMES(s(s(s(x'0'))), 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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 18
                 ↳Forward Instantiation Transformation


Dependency Pairs:

IFTIMES(false, s(s(s(s(x'0')))), y'') -> TIMES(s(s(s(x'0'))), y'')
TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)
IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

IFTIMES(false, s(s(s(s(x'0')))), y'') -> TIMES(s(s(s(x'0'))), y'')
one new Dependency Pair is created:

IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 19
                 ↳Narrowing Transformation


Dependency Pairs:

IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)
TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

TIMES(s(s(s(s(x')))), y) -> IFTIMES(even(x'), s(s(s(s(x')))), y)
three new Dependency Pairs are created:

TIMES(s(s(s(s(0)))), y) -> IFTIMES(true, s(s(s(s(0)))), y)
TIMES(s(s(s(s(s(0))))), y) -> IFTIMES(false, s(s(s(s(s(0))))), y)
TIMES(s(s(s(s(s(s(x'')))))), y) -> IFTIMES(even(x''), s(s(s(s(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 20
                 ↳Narrowing Transformation


Dependency Pairs:

TIMES(s(s(s(s(s(s(x'')))))), y) -> IFTIMES(even(x''), s(s(s(s(s(s(x'')))))), y)
TIMES(s(s(s(s(s(0))))), y) -> IFTIMES(false, s(s(s(s(s(0))))), y)
IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)
TIMES(s(s(s(s(0)))), y) -> IFTIMES(true, s(s(s(s(0)))), y)
IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

IFTIMES(true, s(s(s(s(x')))), y) -> TIMES(s(s(half(x'))), y)
two new Dependency Pairs are created:

IFTIMES(true, s(s(s(s(0)))), y) -> TIMES(s(s(0)), y)
IFTIMES(true, s(s(s(s(s(s(x'')))))), y) -> TIMES(s(s(s(half(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 21
                 ↳Narrowing Transformation


Dependency Pairs:

IFTIMES(true, s(s(s(s(s(s(x'')))))), y) -> TIMES(s(s(s(half(x'')))), y)
TIMES(s(s(s(s(s(0))))), y) -> IFTIMES(false, s(s(s(s(s(0))))), y)
IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
TIMES(s(s(s(s(s(s(x'')))))), y) -> IFTIMES(even(x''), s(s(s(s(s(s(x'')))))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

TIMES(s(s(s(s(s(s(x'')))))), y) -> IFTIMES(even(x''), s(s(s(s(s(s(x'')))))), y)
three new Dependency Pairs are created:

TIMES(s(s(s(s(s(s(0)))))), y) -> IFTIMES(true, s(s(s(s(s(s(0)))))), y)
TIMES(s(s(s(s(s(s(s(0))))))), y) -> IFTIMES(false, s(s(s(s(s(s(s(0))))))), y)
TIMES(s(s(s(s(s(s(s(s(x')))))))), y) -> IFTIMES(even(x'), s(s(s(s(s(s(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 22
                 ↳Narrowing Transformation


Dependency Pairs:

TIMES(s(s(s(s(s(s(s(s(x')))))))), y) -> IFTIMES(even(x'), s(s(s(s(s(s(s(s(x')))))))), y)
TIMES(s(s(s(s(s(s(s(0))))))), y) -> IFTIMES(false, s(s(s(s(s(s(s(0))))))), y)
TIMES(s(s(s(s(s(s(0)))))), y) -> IFTIMES(true, s(s(s(s(s(s(0)))))), y)
IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
TIMES(s(s(s(s(s(0))))), y) -> IFTIMES(false, s(s(s(s(s(0))))), y)
IFTIMES(true, s(s(s(s(s(s(x'')))))), y) -> TIMES(s(s(s(half(x'')))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

IFTIMES(true, s(s(s(s(s(s(x'')))))), y) -> TIMES(s(s(s(half(x'')))), y)
two new Dependency Pairs are created:

IFTIMES(true, s(s(s(s(s(s(0)))))), y) -> TIMES(s(s(s(0))), y)
IFTIMES(true, s(s(s(s(s(s(s(s(x')))))))), y) -> TIMES(s(s(s(s(half(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 23
                 ↳Narrowing Transformation


Dependency Pairs:

IFTIMES(true, s(s(s(s(s(s(s(s(x')))))))), y) -> TIMES(s(s(s(s(half(x'))))), y)
TIMES(s(s(s(s(s(s(s(0))))))), y) -> IFTIMES(false, s(s(s(s(s(s(s(0))))))), y)
TIMES(s(s(s(s(s(0))))), y) -> IFTIMES(false, s(s(s(s(s(0))))), y)
IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
TIMES(s(s(s(s(s(s(s(s(x')))))))), y) -> IFTIMES(even(x'), s(s(s(s(s(s(s(s(x')))))))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

TIMES(s(s(s(s(s(s(s(s(x')))))))), y) -> IFTIMES(even(x'), s(s(s(s(s(s(s(s(x')))))))), y)
three new Dependency Pairs are created:

TIMES(s(s(s(s(s(s(s(s(0)))))))), y) -> IFTIMES(true, s(s(s(s(s(s(s(s(0)))))))), y)
TIMES(s(s(s(s(s(s(s(s(s(0))))))))), y) -> IFTIMES(false, s(s(s(s(s(s(s(s(s(0))))))))), y)
TIMES(s(s(s(s(s(s(s(s(s(s(x'')))))))))), y) -> IFTIMES(even(x''), s(s(s(s(s(s(s(s(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 24
                 ↳Narrowing Transformation


Dependency Pairs:

TIMES(s(s(s(s(s(s(s(s(s(s(x'')))))))))), y) -> IFTIMES(even(x''), s(s(s(s(s(s(s(s(s(s(x'')))))))))), y)
TIMES(s(s(s(s(s(s(s(s(s(0))))))))), y) -> IFTIMES(false, s(s(s(s(s(s(s(s(s(0))))))))), y)
TIMES(s(s(s(s(s(s(s(s(0)))))))), y) -> IFTIMES(true, s(s(s(s(s(s(s(s(0)))))))), y)
TIMES(s(s(s(s(s(s(s(0))))))), y) -> IFTIMES(false, s(s(s(s(s(s(s(0))))))), y)
IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
TIMES(s(s(s(s(s(0))))), y) -> IFTIMES(false, s(s(s(s(s(0))))), y)
IFTIMES(true, s(s(s(s(s(s(s(s(x')))))))), y) -> TIMES(s(s(s(s(half(x'))))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

IFTIMES(true, s(s(s(s(s(s(s(s(x')))))))), y) -> TIMES(s(s(s(s(half(x'))))), y)
two new Dependency Pairs are created:

IFTIMES(true, s(s(s(s(s(s(s(s(0)))))))), y) -> TIMES(s(s(s(s(0)))), y)
IFTIMES(true, s(s(s(s(s(s(s(s(s(s(x'')))))))))), y) -> TIMES(s(s(s(s(s(half(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 25
                 ↳Instantiation Transformation


Dependency Pairs:

IFTIMES(true, s(s(s(s(s(s(s(s(s(s(x'')))))))))), y) -> TIMES(s(s(s(s(s(half(x'')))))), y)
TIMES(s(s(s(s(s(s(s(s(s(0))))))))), y) -> IFTIMES(false, s(s(s(s(s(s(s(s(s(0))))))))), y)
TIMES(s(s(s(s(s(s(s(0))))))), y) -> IFTIMES(false, s(s(s(s(s(s(s(0))))))), y)
TIMES(s(s(s(s(s(0))))), y) -> IFTIMES(false, s(s(s(s(s(0))))), y)
IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
TIMES(s(s(s(s(s(s(s(s(s(s(x'')))))))))), y) -> IFTIMES(even(x''), s(s(s(s(s(s(s(s(s(s(x'')))))))))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

IFTIMES(false, s(s(s(s(s(x'''))))), y''') -> TIMES(s(s(s(s(x''')))), y''')
four new Dependency Pairs are created:

IFTIMES(false, s(s(s(s(s(0))))), y'''') -> TIMES(s(s(s(s(0)))), y'''')
IFTIMES(false, s(s(s(s(s(s(s(0))))))), y'''') -> TIMES(s(s(s(s(s(s(0)))))), y'''')
IFTIMES(false, s(s(s(s(s(s(s(s(s(0))))))))), y'''') -> TIMES(s(s(s(s(s(s(s(s(0)))))))), y'''')
IFTIMES(false, s(s(s(s(s(s(s(s(s(s(x''''')))))))))), y'''') -> TIMES(s(s(s(s(s(s(s(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 26
                 ↳Forward Instantiation Transformation


Dependency Pairs:

IFTIMES(false, s(s(s(s(s(s(s(s(s(s(x''''')))))))))), y'''') -> TIMES(s(s(s(s(s(s(s(s(s(x'''''))))))))), y'''')
TIMES(s(s(s(s(s(s(s(s(s(s(x'')))))))))), y) -> IFTIMES(even(x''), s(s(s(s(s(s(s(s(s(s(x'')))))))))), y)
IFTIMES(true, s(s(s(s(s(s(s(s(s(s(x'')))))))))), y) -> TIMES(s(s(s(s(s(half(x'')))))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost




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

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

IFTIMES(false, s(s(s(s(s(s(s(s(s(s(s(x''''))))))))))), y''''') -> TIMES(s(s(s(s(s(s(s(s(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 4
         ↳Nar
           →DP Problem 14
             ↳Nar
             ...
               →DP Problem 27
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

IFTIMES(false, s(s(s(s(s(s(s(s(s(s(s(x''''))))))))))), y''''') -> TIMES(s(s(s(s(s(s(s(s(s(s(x'''')))))))))), y''''')
IFTIMES(true, s(s(s(s(s(s(s(s(s(s(x'')))))))))), y) -> TIMES(s(s(s(s(s(half(x'')))))), y)
TIMES(s(s(s(s(s(s(s(s(s(s(x'')))))))))), y) -> IFTIMES(even(x''), s(s(s(s(s(s(s(s(s(s(x'')))))))))), y)


Rules:


even(0) -> true
even(s(0)) -> false
even(s(s(x))) -> even(x)
half(0) -> 0
half(s(s(x))) -> s(half(x))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))
times(0, y) -> 0
times(s(x), y) -> iftimes(even(s(x)), s(x), y)
iftimes(true, s(x), y) -> plus(times(half(s(x)), y), times(half(s(x)), y))
iftimes(false, s(x), y) -> plus(y, times(x, y))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:17 minutes