Term Rewriting System R:
[x, y, z, l]
0(#) -> #
+(x, #) -> x
+(#, x) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

+'(0(x), 0(y)) -> 0'(+(x, y))
+'(0(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(1(x), 0(y)) -> +'(x, y)
+'(1(x), 1(y)) -> 0'(+(+(x, y), 1(#)))
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 1(y)) -> +'(x, y)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
*'(0(x), y) -> 0'(*(x, y))
*'(0(x), y) -> *'(x, y)
*'(1(x), y) -> +'(0(*(x, y)), y)
*'(1(x), y) -> 0'(*(x, y))
*'(1(x), y) -> *'(x, y)
*'(*(x, y), z) -> *'(x, *(y, z))
*'(*(x, y), z) -> *'(y, z)
SUM(nil) -> 0'(#)
SUM(cons(x, l)) -> +'(x, sum(l))
SUM(cons(x, l)) -> SUM(l)
PROD(cons(x, l)) -> *'(x, prod(l))
PROD(cons(x, l)) -> PROD(l)

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 Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)

Rules:

0(#) -> #
+(x, #) -> x
+(#, x) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

As we are in the innermost case, we can delete all 8 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`           →DP Problem 5`
`             ↳Modular Removal of Rules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`

Dependency Pairs:

+'(+(x, y), z) -> +'(y, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)
+'(0(x), 0(y)) -> +'(x, y)

Rules:

+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #

Strategy:

innermost

We have the following set of usable rules:

+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(#) =  0 POL(0(x1)) =  x1 POL(1(x1)) =  1 + x1 POL(+(x1, x2)) =  x1 + x2 POL(+'(x1, x2)) =  1 + x1 + x2

We have the following set D of usable symbols: {#, 0, 1, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(1(x), 1(y)) -> +'(x, y)
+'(1(x), 1(y)) -> +'(+(x, y), 1(#))
+'(1(x), 0(y)) -> +'(x, y)
+'(0(x), 1(y)) -> +'(x, y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))

The result of this processor delivers one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`           →DP Problem 5`
`             ↳MRR`
`             ...`
`               →DP Problem 6`
`                 ↳Modular Removal of Rules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`

Dependency Pairs:

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

Rules:

+(x, #) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
0(#) -> #

Strategy:

innermost

We have the following set of usable rules:

+(x, #) -> x
0(#) -> #
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
+(1(x), 0(y)) -> 1(+(x, y))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(#) =  0 POL(0(x1)) =  1 + x1 POL(1(x1)) =  x1 POL(+(x1, x2)) =  x1 + x2 POL(+'(x1, x2)) =  x1 + x2

We have the following set D of usable symbols: {#, 0, 1, +, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

+'(0(x), 0(y)) -> +'(x, y)

The following rules can be deleted as the lhs is strictly greater than the corresponding rhs:

0(#) -> #
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))

The result of this processor delivers one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`           →DP Problem 5`
`             ↳MRR`
`             ...`
`               →DP Problem 7`
`                 ↳Modular Removal of Rules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`

Dependency Pairs:

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

Rules:

+(x, #) -> x
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))

Strategy:

innermost

We have the following set of usable rules:

+(x, #) -> x
+(#, x) -> x
+(+(x, y), z) -> +(x, +(y, z))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(#) =  0 POL(+(x1, x2)) =  x1 + x2 POL(+'(x1, x2)) =  1 + x1 + x2

We have the following set D of usable symbols: {+, +'}
No Dependency Pairs can be deleted.
The following rules can be deleted as they contain symbols in their lhs which do not occur in D:

+(x, #) -> x
+(#, x) -> x

The result of this processor delivers one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`           →DP Problem 5`
`             ↳MRR`
`             ...`
`               →DP Problem 8`
`                 ↳Modular Removal of Rules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`

Dependency Pairs:

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

Rule:

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

Strategy:

innermost

We have the following set of usable rules:

+(+(x, y), z) -> +(x, +(y, z))
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(+(x1, x2)) =  1 + x1 + x2 POL(+'(x1, x2)) =  1 + x1 + x2

We have the following set D of usable symbols: {+, +'}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

No Rules can be deleted.

The result of this processor delivers one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`           →DP Problem 5`
`             ↳MRR`
`             ...`
`               →DP Problem 9`
`                 ↳Dependency Graph`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`

Dependency Pair:

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

Rule:

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

Strategy:

innermost

Using the Dependency Graph resulted in 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 Pairs:

*'(*(x, y), z) -> *'(y, z)
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)

Rules:

0(#) -> #
+(x, #) -> x
+(#, x) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

As we are in the innermost case, we can delete all 16 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 10`
`             ↳Size-Change Principle`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`

Dependency Pairs:

*'(*(x, y), z) -> *'(y, z)
*'(1(x), y) -> *'(x, y)
*'(0(x), y) -> *'(x, y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. *'(*(x, y), z) -> *'(y, z)
2. *'(1(x), y) -> *'(x, y)
3. *'(0(x), y) -> *'(x, y)
and get the following Size-Change Graph(s):
{1, 2, 3} , {1, 2, 3}
1>1
2=2

which lead(s) to this/these maximal multigraph(s):
{1, 2, 3} , {1, 2, 3}
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:
0(x1) -> 0(x1)
1(x1) -> 1(x1)
*(x1, x2) -> *(x1, x2)

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:

SUM(cons(x, l)) -> SUM(l)

Rules:

0(#) -> #
+(x, #) -> x
+(#, x) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

As we are in the innermost case, we can delete all 16 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`           →DP Problem 11`
`             ↳Size-Change Principle`
`       →DP Problem 4`
`         ↳UsableRules`

Dependency Pair:

SUM(cons(x, l)) -> SUM(l)

Rule:

none

Strategy:

innermost

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

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

DP: empty set
Oriented Rules: none

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

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

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:

PROD(cons(x, l)) -> PROD(l)

Rules:

0(#) -> #
+(x, #) -> x
+(#, x) -> x
+(0(x), 0(y)) -> 0(+(x, y))
+(0(x), 1(y)) -> 1(+(x, y))
+(1(x), 0(y)) -> 1(+(x, y))
+(1(x), 1(y)) -> 0(+(+(x, y), 1(#)))
+(+(x, y), z) -> +(x, +(y, z))
*(#, x) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
*(*(x, y), z) -> *(x, *(y, z))
sum(nil) -> 0(#)
sum(cons(x, l)) -> +(x, sum(l))
prod(nil) -> 1(#)
prod(cons(x, l)) -> *(x, prod(l))

Strategy:

innermost

As we are in the innermost case, we can delete all 16 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`       →DP Problem 4`
`         ↳UsableRules`
`           →DP Problem 12`
`             ↳Size-Change Principle`

Dependency Pair:

PROD(cons(x, l)) -> PROD(l)

Rule:

none

Strategy:

innermost

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

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

DP: empty set
Oriented Rules: none

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

with Argument Filtering System:
cons(x1, x2) -> cons(x1, x2)

We obtain no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes