Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(x, g(y, z)) -> F(x, y)
++'(x, g(y, z)) -> ++'(x, y)
MEM(g(x, y), z) -> MEM(x, z)
MEM(x, max(x)) -> NULL(x)
MAX(g(g(g(x, y), z), u)) -> MAX(g(g(x, y), z))

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

F(x, g(y, z)) -> F(x, y)

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

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

F(x, g(y, z)) -> F(x, y)

one new Dependency Pair is created:

F(x'', g(g(y'', z''), z)) -> F(x'', g(y'', z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

F(x'', g(g(y'', z''), z)) -> F(x'', g(y'', z''))

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

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

F(x'', g(g(y'', z''), z)) -> F(x'', g(y'', z''))

one new Dependency Pair is created:

F(x'''', g(g(g(y'''', z''''), z''0), z)) -> F(x'''', g(g(y'''', z''''), z''0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 6

                 ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

F(x'''', g(g(g(y'''', z''''), z''0), z)) -> F(x'''', g(g(y'''', z''''), z''0))

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(x'''', g(g(g(y'''', z''''), z''0), z)) -> F(x'''', g(g(y'''', z''''), z''0))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

F(x₁, x₂) -> F(x₁, x₂)
g(x₁, x₂) -> g(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

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

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

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

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

one new Dependency Pair is created:

++'(x'''', g(g(g(y'''', z''''), z''0), z)) -> ++'(x'''', g(g(y'''', z''''), z''0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

++'(x'''', g(g(g(y'''', z''''), z''0), z)) -> ++'(x'''', g(g(y'''', z''''), z''0))

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

The following dependency pair can be strictly oriented:

++'(x'''', g(g(g(y'''', z''''), z''0), z)) -> ++'(x'''', g(g(y'''', z''''), z''0))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 8

             ↳FwdInst

...

               →DP Problem 10

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

MEM(g(x, y), z) -> MEM(x, z)

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

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

MEM(g(x, y), z) -> MEM(x, z)

one new Dependency Pair is created:

MEM(g(g(x'', y''), y), z'') -> MEM(g(x'', y''), z'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳Forward Instantiation Transformation

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

MEM(g(g(x'', y''), y), z'') -> MEM(g(x'', y''), z'')

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

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

MEM(g(g(x'', y''), y), z'') -> MEM(g(x'', y''), z'')

one new Dependency Pair is created:

MEM(g(g(g(x'''', y''''), y''0), y), z'''') -> MEM(g(g(x'''', y''''), y''0), z'''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 12

                 ↳Argument Filtering and Ordering

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

MEM(g(g(g(x'''', y''''), y''0), y), z'''') -> MEM(g(g(x'''', y''''), y''0), z'''')

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

The following dependency pair can be strictly oriented:

MEM(g(g(g(x'''', y''''), y''0), y), z'''') -> MEM(g(g(x'''', y''''), y''0), z'''')

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

MEM(x₁, x₂) -> MEM(x₁, x₂)
g(x₁, x₂) -> g(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 11

             ↳FwdInst

...

               →DP Problem 13

                 ↳Dependency Graph

       →DP Problem 4

         ↳FwdInst

Dependency Pair:

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳Forward Instantiation Transformation

Dependency Pair:

MAX(g(g(g(x, y), z), u)) -> MAX(g(g(x, y), z))

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

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

MAX(g(g(g(x, y), z), u)) -> MAX(g(g(x, y), z))

one new Dependency Pair is created:

MAX(g(g(g(g(x'', y''), y0), u), u)) -> MAX(g(g(g(x'', y''), y0), u))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳Forward Instantiation Transformation

Dependency Pair:

MAX(g(g(g(g(x'', y''), y0), u), u)) -> MAX(g(g(g(x'', y''), y0), u))

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

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

MAX(g(g(g(g(x'', y''), y0), u), u)) -> MAX(g(g(g(x'', y''), y0), u))

one new Dependency Pair is created:

MAX(g(g(g(g(g(x'''', y''''), y''0), u), u), u)) -> MAX(g(g(g(g(x'''', y''''), y''0), u), u))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 15

                 ↳Argument Filtering and Ordering

Dependency Pair:

MAX(g(g(g(g(g(x'''', y''''), y''0), u), u), u)) -> MAX(g(g(g(g(x'''', y''''), y''0), u), u))

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

The following dependency pair can be strictly oriented:

MAX(g(g(g(g(g(x'''', y''''), y''0), u), u), u)) -> MAX(g(g(g(g(x'''', y''''), y''0), u), u))

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

resulting in one new DP problem.
Used Argument Filtering System:

MAX(x₁) -> MAX(x₁)
g(x₁, x₂) -> g(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

       →DP Problem 4

         ↳FwdInst

           →DP Problem 14

             ↳FwdInst

...

               →DP Problem 16

                 ↳Dependency Graph

Dependency Pair:

Rules:

f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
++(x, nil) -> x
++(x, g(y, z)) -> g(++(x, y), z)
null(nil) -> true
null(g(x, y)) -> false
mem(nil, y) -> false
mem(g(x, y), z) -> or(=(y, z), mem(x, z))
mem(x, max(x)) -> not(null(x))
max(g(g(nil, x), y)) -> max'(x, y)
max(g(g(g(x, y), z), u)) -> max'(max(g(g(x, y), z)), u)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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