Term Rewriting System R:
[x, y, 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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
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)
*'(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)
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:

+'(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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
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 7 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:

+'(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
+(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
+(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 Pair:

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

Strategy:

innermost

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(0(x1)) =  x1 POL(+'(x1, x2)) =  x1 + x2

We have the following set D of usable symbols: {+'}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

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

Dependency Pairs:

*'(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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
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 14 non-usable-rules.

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

Dependency Pairs:

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

Rule:

none

Strategy:

innermost

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

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

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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
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 14 non-usable-rules.

R
DPs
→DP Problem 1
UsableRules
→DP Problem 2
UsableRules
→DP Problem 3
UsableRules
→DP Problem 8
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) -> #
*(0(x), y) -> 0(*(x, y))
*(1(x), y) -> +(0(*(x, y)), y)
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 14 non-usable-rules.

R
DPs
→DP Problem 1
UsableRules
→DP Problem 2
UsableRules
→DP Problem 3
UsableRules
→DP Problem 4
UsableRules
→DP Problem 9
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