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))

Innermost 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
Argument Filtering and Ordering
→DP Problem 2
AFS

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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(++'(x1, x2)) =  x1 + x2 POL(.(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.
Used Argument Filtering System:
++'(x1, x2) -> ++'(x1, x2)
.(x1, x2) -> .(x1, x2)

R
DPs
→DP Problem 1
AFS
→DP Problem 3
Dependency Graph
→DP Problem 2
AFS

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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
AFS
→DP Problem 2
Argument Filtering and 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))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(REV(x1)) =  x1 POL(.(x1, x2)) =  1 + x1 + x2

resulting in one new DP problem.
Used Argument Filtering System:
REV(x1) -> REV(x1)
.(x1, x2) -> .(x1, x2)

R
DPs
→DP Problem 1
AFS
→DP Problem 2
AFS
→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))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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