Term Rewriting System R:
[x, y, z]
rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))

Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

REV(.(x, y)) -> ++'(rev(y), .(x, nil))
REV(.(x, y)) -> REV(y)
++'(.(x, y), z) -> ++'(y, z)

Furthermore, R contains two SCCs.

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

Dependency Pair:

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

Rules:

rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))

The following dependency pair can be strictly oriented:

++'(.(x, y), z) -> ++'(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(++'(x1, x2)) =  x1 POL(.(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`           →DP Problem 3`
`             ↳Dependency Graph`
`       →DP Problem 2`
`         ↳Polo`

Dependency Pair:

Rules:

rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

REV(.(x, y)) -> REV(y)

Rules:

rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))

The following dependency pair can be strictly oriented:

REV(.(x, y)) -> REV(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(REV(x1)) =  x1 POL(.(x1, x2)) =  1 + x2

resulting in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Polo`
`       →DP Problem 2`
`         ↳Polo`
`           →DP Problem 4`
`             ↳Dependency Graph`

Dependency Pair:

Rules:

rev(nil) -> nil
rev(.(x, y)) -> ++(rev(y), .(x, nil))
car(.(x, y)) -> x
cdr(.(x, y)) -> y
null(nil) -> true
null(.(x, y)) -> false
++(nil, y) -> y
++(.(x, y), z) -> .(x, ++(y, z))

Using the Dependency Graph resulted in no new DP problems.

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