Termination Proof using AProVE

Term Rewriting System R:
[z, x, y]
+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(y, z))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

+'(a, b) -> +'(b, a)
+'(a, +(b, z)) -> +'(b, +(a, z))
+'(a, +(b, z)) -> +'(a, z)
+'(+(x, y), z) -> +'(x, +(y, z))
+'(+(x, y), z) -> +'(y, z)
F(+(x, y), z) -> +'(f(x, z), f(y, z))
F(+(x, y), z) -> F(x, z)
F(+(x, y), z) -> F(y, z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

+'(a, +(b, z)) -> +'(a, z)

Rules:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(y, z))

Strategy:

innermost

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

+'(a, +(b, z)) -> +'(a, z)

one new Dependency Pair is created:

+'(a, +(b, +(b, z''))) -> +'(a, +(b, z''))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

+'(a, +(b, +(b, z''))) -> +'(a, +(b, z''))

Rules:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

+'(a, +(b, +(b, z''))) -> +'(a, +(b, z''))

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(b) = 0

POL(a) = 0

POL(+(x₁, x₂)) = 1 + x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Polo

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(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

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(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 2

         ↳FwdInst

           →DP Problem 6

             ↳Polynomial Ordering

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

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

Rules:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(b) = 0

POL(a) = 0

POL(+(x₁, x₂)) = 1 + x₁ + x₂

POL(+'(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 6

             ↳Polo

...

               →DP Problem 7

                 ↳Dependency Graph

       →DP Problem 3

         ↳FwdInst

Dependency Pair:

Rules:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(y, z))

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

Dependency Pair:

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

Rules:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(y, z))

Strategy:

innermost

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

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

one new Dependency Pair is created:

F(+(x, +(x'', y'')), z'') -> F(+(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 8

             ↳Polynomial Ordering

Dependency Pair:

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

Rules:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(y, z))

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost w.r.t. to the implicit AFS can be oriented:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(b) = 0

POL(a) = 0

POL(+(x₁, x₂)) = 1 + x₁ + x₂

POL(F(x₁, x₂)) = 1 + x₁ + x₂

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳FwdInst

           →DP Problem 8

             ↳Polo

...

               →DP Problem 9

                 ↳Dependency Graph

Dependency Pair:

Rules:

+(a, b) -> +(b, a)
+(a, +(b, z)) -> +(b, +(a, z))
+(+(x, y), z) -> +(x, +(y, z))
f(a, y) -> a
f(b, y) -> b
f(+(x, y), z) -> +(f(x, z), f(y, z))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

POL(b)	= 0
POL(a)	= 0
POL(+(x₁, x₂))	= 1 + x₁ + x₂
POL(+'(x₁, x₂))	= 1 + x₁ + x₂