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
Forward Instantiation Transformation
→DP Problem 2
FwdInst

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

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

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

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

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:

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

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

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

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

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:

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

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(.(x1, x2)) =  1 + x2

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:

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
FwdInst
→DP Problem 2
Forward Instantiation Transformation

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

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

REV(.(x, y)) -> REV(y)
one new Dependency Pair is created:

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

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

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

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

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

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

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:

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