Termination Proof using AProVE

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.

     ↳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.

     ↳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:

     ↳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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Argument Filtering and 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 AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

++'(x₁, x₂) -> ++'(x₁, x₂)
.(x₁, x₂) -> .(x₁, x₂)

     ↳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.

     ↳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:

     ↳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:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Argument Filtering and 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'''')))

REV(x₁) -> REV(x₁)
.(x₁, x₂) -> .(x₁, x₂)

     ↳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