Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

F(s(0), y, z) -> F(0, s(y), s(z))
F(s(x), s(y), 0) -> F(x, y, s(0))
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(0, s(s(y)), s(0)) -> F(0, y, s(0))
F(0, s(0), s(s(z))) -> F(0, s(0), z)
F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))
F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))

Furthermore, R contains four SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

F(0, s(s(y)), s(0)) -> F(0, y, s(0))

Rules:

f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(0, s(s(y)), s(0)) -> F(0, y, s(0))

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(s(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃

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

F(x₁, x₂, x₃) -> F(x₁, x₂, x₃)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(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

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

F(0, s(0), s(s(z))) -> F(0, s(0), z)

Rules:

f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Strategy:

innermost

The following dependency pair can be strictly oriented:

F(0, s(0), s(s(z))) -> F(0, s(0), z)

There are no usable rules for innermost that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(0) = 0

POL(s(x₁)) = 1 + x₁

POL(F(x₁, x₂, x₃)) = 1 + x₁ + x₂ + x₃

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

F(x₁, x₂, x₃) -> F(x₁, x₂, x₃)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 6

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining

Dependency Pair:

Rules:

f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(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

         ↳Narrowing Transformation

       →DP Problem 4

         ↳Remaining

Dependency Pairs:

F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))
F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))

Rules:

f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Strategy:

innermost

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

F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))

two new Dependency Pairs are created:

F(0, s(s(y'')), s(s(0))) -> F(0, y'', f(0, y'', s(0)))
F(0, s(s(y'')), s(s(s(z'')))) -> F(0, y'', f(0, y'', f(0, s(s(y'')), s(z''))))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
F(0, s(s(y'')), s(s(s(z'')))) -> F(0, y'', f(0, y'', f(0, s(s(y'')), s(z''))))
F(0, s(s(y'')), s(s(0))) -> F(0, y'', f(0, y'', s(0)))
F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))

Rules:

f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Strategy:
innermost
Dependency Pairs:
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), 0) -> F(x, y, s(0))

Rules:

f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Strategy:
innermost

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

       →DP Problem 4

         ↳Remaining Obligation(s)

The following remains to be proven:

Dependency Pairs:
F(0, s(s(y'')), s(s(s(z'')))) -> F(0, y'', f(0, y'', f(0, s(s(y'')), s(z''))))
F(0, s(s(y'')), s(s(0))) -> F(0, y'', f(0, y'', s(0)))
F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))

Rules:

f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Strategy:
innermost
Dependency Pairs:
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), 0) -> F(x, y, s(0))

Rules:

f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))

Strategy:
innermost

Innermost Termination of R could not be shown.
Duration:
0:02 minutes

POL(0)	= 0
POL(s(x₁))	= 1 + x₁
POL(F(x₁, x₂, x₃))	= 1 + x₁ + x₂ + x₃