Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MAX(N(L(s(x)), L(s(y)))) -> MAX(N(L(x), L(y)))
MAX(N(L(x), N(y, z))) -> MAX(N(L(x), L(max(N(y, z)))))
MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

MAX(N(L(s(x)), L(s(y)))) -> MAX(N(L(x), L(y)))

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

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

MAX(N(L(s(x)), L(s(y)))) -> MAX(N(L(x), L(y)))

one new Dependency Pair is created:

MAX(N(L(s(s(x''))), L(s(s(y''))))) -> MAX(N(L(s(x'')), L(s(y''))))

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:

MAX(N(L(s(s(x''))), L(s(s(y''))))) -> MAX(N(L(s(x'')), L(s(y''))))

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

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

MAX(N(L(s(s(x''))), L(s(s(y''))))) -> MAX(N(L(s(x'')), L(s(y''))))

one new Dependency Pair is created:

MAX(N(L(s(s(s(x'''')))), L(s(s(s(y'''')))))) -> MAX(N(L(s(s(x''''))), L(s(s(y'''')))))

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:

MAX(N(L(s(s(s(x'''')))), L(s(s(s(y'''')))))) -> MAX(N(L(s(s(x''''))), L(s(s(y'''')))))

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MAX(N(L(s(s(s(x'''')))), L(s(s(s(y'''')))))) -> MAX(N(L(s(s(x''''))), L(s(s(y'''')))))

There are no usable rules for innermost 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)
N(x₁, x₂) -> N(x₁, x₂)
L(x₁) -> L(x₁)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(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:

MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

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

MAX(N(L(x), N(y, z))) -> MAX(N(y, z))

one new Dependency Pair is created:

MAX(N(L(x), N(L(x''), N(y'', z'')))) -> MAX(N(L(x''), N(y'', z'')))

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:

MAX(N(L(x), N(L(x''), N(y'', z'')))) -> MAX(N(L(x''), N(y'', z'')))

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

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

MAX(N(L(x), N(L(x''), N(y'', z'')))) -> MAX(N(L(x''), N(y'', z'')))

one new Dependency Pair is created:

MAX(N(L(x), N(L(x''''), N(L(x'''''), N(y'''', z''''))))) -> MAX(N(L(x''''), N(L(x'''''), N(y'''', z''''))))

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:

MAX(N(L(x), N(L(x''''), N(L(x'''''), N(y'''', z''''))))) -> MAX(N(L(x''''), N(L(x'''''), N(y'''', z''''))))

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

The following dependency pair can be strictly oriented:

MAX(N(L(x), N(L(x''''), N(L(x'''''), N(y'''', z''''))))) -> MAX(N(L(x''''), N(L(x'''''), N(y'''', z''''))))

There are no usable rules for innermost 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)
N(x₁, x₂) -> N(x₁, x₂)
L(x₁) -> L(x₁)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

max(L(x)) -> x
max(N(L(0), L(y))) -> y
max(N(L(s(x)), L(s(y)))) -> s(max(N(L(x), L(y))))
max(N(L(x), N(y, z))) -> max(N(L(x), L(max(N(y, z)))))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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