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

Innermost Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

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

Furthermore, R contains two SCCs.

R
DPs
→DP Problem 1
Forward Instantiation Transformation
→DP Problem 2
FwdInst

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
FwdInst
→DP Problem 3
Forward Instantiation Transformation
→DP Problem 2
FwdInst

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
FwdInst
→DP Problem 3
FwdInst
...
→DP Problem 4
Polynomial Ordering
→DP Problem 2
FwdInst

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(s(x1)) =  1 + x1 POL(+'(x1, x2)) =  1 + x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
FwdInst
→DP Problem 3
FwdInst
...
→DP Problem 5
Dependency Graph
→DP Problem 2
FwdInst

Dependency Pair:

Rules:

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

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
FwdInst
→DP Problem 2
FwdInst
→DP Problem 6
Forward Instantiation Transformation

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

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

The transformation is resulting in one new DP problem:

R
DPs
→DP Problem 1
FwdInst
→DP Problem 2
FwdInst
→DP Problem 6
FwdInst
...
→DP Problem 7
Polynomial Ordering

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(-'(x1, x2)) =  1 + x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
FwdInst
→DP Problem 2
FwdInst
→DP Problem 6
FwdInst
...
→DP Problem 8
Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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