Term Rewriting System R:
[x, y, z, l, l1, l2]
+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

+'(s(x), s(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
*'(s(x), s(y)) -> +'(*(x, y), +(x, y))
*'(s(x), s(y)) -> *'(x, y)
*'(s(x), s(y)) -> +'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)
APP(cons(x, l1), l2) -> APP(l1, l2)
SUM(cons(x, l)) -> +'(x, sum(l))
SUM(cons(x, l)) -> SUM(l)
SUM(app(l1, l2)) -> +'(sum(l1), sum(l2))
SUM(app(l1, l2)) -> SUM(l1)
SUM(app(l1, l2)) -> SUM(l2)
PROD(cons(x, l)) -> *'(x, prod(l))
PROD(cons(x, l)) -> PROD(l)
PROD(app(l1, l2)) -> *'(prod(l1), prod(l2))
PROD(app(l1, l2)) -> PROD(l1)
PROD(app(l1, l2)) -> PROD(l2)

Furthermore, R contains five SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`

Dependency Pairs:

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

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

The following dependency pairs can be strictly oriented:

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

The following usable rules using the Ce-refinement can be oriented:

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
{+', +} > s

resulting in one new DP problem.
Used Argument Filtering System:
+'(x1, x2) -> +'(x1, x2)
s(x1) -> s(x1)
+(x1, x2) -> +(x1, x2)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`           →DP Problem 6`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

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`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`

Dependency Pair:

APP(cons(x, l1), l2) -> APP(l1, l2)

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

The following dependency pair can be strictly oriented:

APP(cons(x, l1), l2) -> APP(l1, l2)

There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
APP(x1, x2) -> APP(x1, x2)
cons(x1, x2) -> cons(x1, x2)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`           →DP Problem 7`
`             ↳Dependency Graph`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`

Dependency Pairs:

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

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

The following dependency pairs can be strictly oriented:

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

The following usable rules using the Ce-refinement can be oriented:

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

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
*' > * > + > s

resulting in one new DP problem.
Used Argument Filtering System:
*'(x1, x2) -> *'(x1, x2)
s(x1) -> s(x1)
*(x1, x2) -> *(x1, x2)
+(x1, x2) -> +(x1, x2)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`           →DP Problem 8`
`             ↳Dependency Graph`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳Argument Filtering and Ordering`
`       →DP Problem 5`
`         ↳AFS`

Dependency Pairs:

SUM(app(l1, l2)) -> SUM(l2)
SUM(app(l1, l2)) -> SUM(l1)
SUM(cons(x, l)) -> SUM(l)

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

The following dependency pairs can be strictly oriented:

SUM(app(l1, l2)) -> SUM(l2)
SUM(app(l1, l2)) -> SUM(l1)
SUM(cons(x, l)) -> SUM(l)

There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
SUM(x1) -> SUM(x1)
app(x1, x2) -> app(x1, x2)
cons(x1, x2) -> cons(x1, x2)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`           →DP Problem 9`
`             ↳Dependency Graph`
`       →DP Problem 5`
`         ↳AFS`

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳Argument Filtering and Ordering`

Dependency Pairs:

PROD(app(l1, l2)) -> PROD(l2)
PROD(app(l1, l2)) -> PROD(l1)
PROD(cons(x, l)) -> PROD(l)

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

The following dependency pairs can be strictly oriented:

PROD(app(l1, l2)) -> PROD(l2)
PROD(app(l1, l2)) -> PROD(l1)
PROD(cons(x, l)) -> PROD(l)

There are no usable rules using the Ce-refinement that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:
trivial

resulting in one new DP problem.
Used Argument Filtering System:
PROD(x1) -> PROD(x1)
cons(x1, x2) -> cons(x1, x2)
app(x1, x2) -> app(x1, x2)

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳AFS`
`       →DP Problem 2`
`         ↳AFS`
`       →DP Problem 3`
`         ↳AFS`
`       →DP Problem 4`
`         ↳AFS`
`       →DP Problem 5`
`         ↳AFS`
`           →DP Problem 10`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

+(x, 0) -> x
+(0, x) -> x
+(s(x), s(y)) -> s(s(+(x, y)))
+(+(x, y), z) -> +(x, +(y, z))
*(x, 0) -> 0
*(0, x) -> 0
*(s(x), s(y)) -> s(+(*(x, y), +(x, y)))
*(*(x, y), z) -> *(x, *(y, z))
app(nil, l) -> l
app(cons(x, l1), l2) -> cons(x, app(l1, l2))
sum(nil) -> 0
sum(cons(x, l)) -> +(x, sum(l))
sum(app(l1, l2)) -> +(sum(l1), sum(l2))
prod(nil) -> s(0)
prod(cons(x, l)) -> *(x, prod(l))
prod(app(l1, l2)) -> *(prod(l1), prod(l2))

Using the Dependency Graph resulted in no new DP problems.

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