Term Rewriting System R:
[x, y, z]
minus(0) -> 0
minus(minus(x)) -> x
+(x, 0) -> x
+(0, y) -> y
+(minus(1), 1) -> 0
+(x, minus(y)) -> minus(+(minus(x), y))
+(x, +(y, z)) -> +(+(x, y), z)
+(minus(+(x, 1)), 1) -> minus(x)

Termination of R to be shown.

R
Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(x, minus(y)) -> MINUS(+(minus(x), y))
+'(x, minus(y)) -> +'(minus(x), y)
+'(x, minus(y)) -> MINUS(x)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, +(y, z)) -> +'(x, y)
+'(minus(+(x, 1)), 1) -> MINUS(x)

Furthermore, R contains one SCC.

R
DPs
→DP Problem 1
Polynomial Ordering

Dependency Pairs:

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

Rules:

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

The following dependency pairs can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(1) =  0 POL(minus(x1)) =  x1 POL(+(x1, x2)) =  1 + x1 + x2 POL(+'(x1, x2)) =  x2

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polynomial Ordering

Dependency Pair:

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

Rules:

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

The following dependency pair can be strictly oriented:

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

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(minus(x1)) =  1 + x1 POL(+'(x1, x2)) =  x2

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
...
→DP Problem 3
Dependency Graph

Dependency Pair:

Rules:

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

Using the Dependency Graph resulted in no new DP problems.

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