Termination Proof using AProVE

Term Rewriting System R:
[x, y, z]
norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

NORM(g(x, y)) -> NORM(x)
F(x, g(y, z)) -> F(x, y)
REM(g(x, y), s(z)) -> REM(x, z)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

NORM(g(x, y)) -> NORM(x)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

NORM(g(x, y)) -> NORM(x)

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Homeomorphic Embedding Order with EMB
resulting in one new DP problem.
Used Argument Filtering System:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, 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:

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

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳AFS

Dependency Pair:

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, 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:

REM(g(x, y), s(z)) -> REM(x, z)

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

REM(g(x, y), s(z)) -> REM(x, z)

REM(x₁, x₂) -> REM(x₁, x₂)
g(x₁, x₂) -> g(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

Dependency Pair:

Rules:

norm(nil) -> 0
norm(g(x, y)) -> s(norm(x))
f(x, nil) -> g(nil, x)
f(x, g(y, z)) -> g(f(x, y), z)
rem(nil, y) -> nil
rem(g(x, y), 0) -> g(x, y)
rem(g(x, y), s(z)) -> rem(x, z)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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