Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

QUOT(s(x), s(y), z) -> QUOT(x, y, z)
QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
QUOT(x, 0, s(z)) -> PLUS(z, s(0))
PLUS(s(x), y) -> PLUS(x, y)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳Nar

Dependency Pair:

PLUS(s(x), y) -> PLUS(x, y)

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

PLUS(s(x), y) -> PLUS(x, y)

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

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Nar

Dependency Pair:

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Narrowing Transformation

Dependency Pairs:

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

QUOT(x, 0, s(z)) -> QUOT(x, plus(z, s(0)), s(z))

two new Dependency Pairs are created:

QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳Forward Instantiation Transformation

Dependency Pairs:

QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))
QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(s(x), s(y), z) -> QUOT(x, y, z)

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

QUOT(s(x), s(y), z) -> QUOT(x, y, z)

three new Dependency Pairs are created:

QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))
QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Forward Instantiation Transformation

Dependency Pairs:

QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))
QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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

QUOT(x, 0, s(0)) -> QUOT(x, s(0), s(0))

one new Dependency Pair is created:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

Dependency Pairs:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))
QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

The following dependency pair can be strictly oriented:

QUOT(s(x''), s(0), s(0)) -> QUOT(x'', 0, s(0))

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

QUOT(x₁, x₂, x₃) -> x₁
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

QUOT(s(x''''), 0, s(0)) -> QUOT(s(x''''), s(0), s(0))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Nar

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

QUOT(s(x''), s(0), s(s(x''''))) -> QUOT(x'', 0, s(s(x'''')))
QUOT(s(s(x'')), s(s(y'')), z'') -> QUOT(s(x''), s(y''), z'')
QUOT(x, 0, s(s(x''))) -> QUOT(x, s(plus(x'', s(0))), s(s(x'')))

Rules:

quot(0, s(y), s(z)) -> 0
quot(s(x), s(y), z) -> quot(x, y, z)
quot(x, 0, s(z)) -> s(quot(x, plus(z, s(0)), s(z)))
plus(0, y) -> y
plus(s(x), y) -> s(plus(x, y))

Strategy:

innermost

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