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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

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

The following dependency pair can be strictly oriented:

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

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

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳AFS

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

The following dependency pair can be strictly oriented:

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

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

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Argument Filtering and Ordering

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

The following dependency pair can be strictly oriented:

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳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