Termination Proof using AProVE

Term Rewriting System R:
[X, Y, Z]
plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

PLUS(plus(X, Y), Z) -> PLUS(X, plus(Y, Z))
PLUS(plus(X, Y), Z) -> PLUS(Y, Z)
TIMES(X, s(Y)) -> PLUS(X, times(Y, X))
TIMES(X, s(Y)) -> TIMES(Y, X)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

PLUS(plus(X, Y), Z) -> PLUS(Y, Z)

Rules:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Strategy:

innermost

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

PLUS(plus(X, Y), Z) -> PLUS(Y, Z)

one new Dependency Pair is created:

PLUS(plus(X, plus(X'', Y'')), Z'') -> PLUS(plus(X'', Y''), Z'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

PLUS(plus(X, plus(X'', Y'')), Z'') -> PLUS(plus(X'', Y''), Z'')

Rules:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Strategy:

innermost

The following dependency pair can be strictly oriented:

PLUS(plus(X, plus(X'', Y'')), Z'') -> PLUS(plus(X'', Y''), Z'')

The following usable rule for innermost w.r.t. to the AFS can be oriented:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))

Used ordering: Lexicographic Path Order with Non-Strict Precedence with Quasi Precedence:

trivial

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

PLUS(x₁, x₂) -> PLUS(x₁, x₂)
plus(x₁, x₂) -> plus(x₁, x₂)

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳AFS

...

               →DP Problem 4

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

Dependency Pair:

Rules:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

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:

TIMES(X, s(Y)) -> TIMES(Y, X)

Rules:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Strategy:

innermost

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

TIMES(X, s(Y)) -> TIMES(Y, X)

one new Dependency Pair is created:

TIMES(s(Y''), s(Y0)) -> TIMES(Y0, s(Y''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳Forward Instantiation Transformation

Dependency Pair:

TIMES(s(Y''), s(Y0)) -> TIMES(Y0, s(Y''))

Rules:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Strategy:

innermost

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

TIMES(s(Y''), s(Y0)) -> TIMES(Y0, s(Y''))

one new Dependency Pair is created:

TIMES(s(Y''''), s(s(Y'''''))) -> TIMES(s(Y'''''), s(Y''''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 6

                 ↳Forward Instantiation Transformation

Dependency Pair:

TIMES(s(Y''''), s(s(Y'''''))) -> TIMES(s(Y'''''), s(Y''''))

Rules:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Strategy:

innermost

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

TIMES(s(Y''''), s(s(Y'''''))) -> TIMES(s(Y'''''), s(Y''''))

one new Dependency Pair is created:

TIMES(s(s(Y'''''''')), s(s(Y'''''0))) -> TIMES(s(Y'''''0), s(s(Y'''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 7

                 ↳Forward Instantiation Transformation

Dependency Pair:

TIMES(s(s(Y'''''''')), s(s(Y'''''0))) -> TIMES(s(Y'''''0), s(s(Y'''''''')))

Rules:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Strategy:

innermost

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

TIMES(s(s(Y'''''''')), s(s(Y'''''0))) -> TIMES(s(Y'''''0), s(s(Y'''''''')))

one new Dependency Pair is created:

TIMES(s(s(Y'''''''''')), s(s(s(Y''''''''''')))) -> TIMES(s(s(Y''''''''''')), s(s(Y'''''''''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 8

                 ↳Forward Instantiation Transformation

Dependency Pair:

TIMES(s(s(Y'''''''''')), s(s(s(Y''''''''''')))) -> TIMES(s(s(Y''''''''''')), s(s(Y'''''''''')))

Rules:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Strategy:

innermost

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

TIMES(s(s(Y'''''''''')), s(s(s(Y''''''''''')))) -> TIMES(s(s(Y''''''''''')), s(s(Y'''''''''')))

one new Dependency Pair is created:

TIMES(s(s(s(Y''''''''''''''))), s(s(s(Y'''''''''''0)))) -> TIMES(s(s(Y'''''''''''0)), s(s(s(Y''''''''''''''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 5

             ↳FwdInst

...

               →DP Problem 9

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

TIMES(s(s(s(Y''''''''''''''))), s(s(s(Y'''''''''''0)))) -> TIMES(s(s(Y'''''''''''0)), s(s(s(Y''''''''''''''))))

Rules:

plus(plus(X, Y), Z) -> plus(X, plus(Y, Z))
times(X, s(Y)) -> plus(X, times(Y, X))

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes