Termination Proof using AProVE

Term Rewriting System R:
[y, x, z]
minus_active(0, y) -> 0
minus_active(s(x), s(y)) -> minus_active(x, y)
minus_active(x, y) -> minus(x, y)
mark(0) -> 0
mark(s(x)) -> s(mark(x))
mark(minus(x, y)) -> minus_active(x, y)
mark(ge(x, y)) -> ge_active(x, y)
mark(div(x, y)) -> div_active(mark(x), y)
mark(if(x, y, z)) -> if_active(mark(x), y, z)
ge_active(x, 0) -> true
ge_active(0, s(y)) -> false
ge_active(s(x), s(y)) -> ge_active(x, y)
ge_active(x, y) -> ge(x, y)
div_active(0, s(y)) -> 0
div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
div_active(x, y) -> div(x, y)
if_active(true, x, y) -> mark(x)
if_active(false, x, y) -> mark(y)
if_active(x, y, z) -> if(x, y, z)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS_ACTIVE(s(x), s(y)) -> MINUS_ACTIVE(x, y)
MARK(s(x)) -> MARK(x)
MARK(minus(x, y)) -> MINUS_ACTIVE(x, y)
MARK(ge(x, y)) -> GE_ACTIVE(x, y)
MARK(div(x, y)) -> DIV_ACTIVE(mark(x), y)
MARK(div(x, y)) -> MARK(x)
MARK(if(x, y, z)) -> IF_ACTIVE(mark(x), y, z)
MARK(if(x, y, z)) -> MARK(x)
GE_ACTIVE(s(x), s(y)) -> GE_ACTIVE(x, y)
DIV_ACTIVE(s(x), s(y)) -> IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
DIV_ACTIVE(s(x), s(y)) -> GE_ACTIVE(x, y)
IF_ACTIVE(true, x, y) -> MARK(x)
IF_ACTIVE(false, x, y) -> MARK(y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Rules:

minus_active(0, y) -> 0
minus_active(s(x), s(y)) -> minus_active(x, y)
minus_active(x, y) -> minus(x, y)
mark(0) -> 0
mark(s(x)) -> s(mark(x))
mark(minus(x, y)) -> minus_active(x, y)
mark(ge(x, y)) -> ge_active(x, y)
mark(div(x, y)) -> div_active(mark(x), y)
mark(if(x, y, z)) -> if_active(mark(x), y, z)
ge_active(x, 0) -> true
ge_active(0, s(y)) -> false
ge_active(s(x), s(y)) -> ge_active(x, y)
ge_active(x, y) -> ge(x, y)
div_active(0, s(y)) -> 0
div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
div_active(x, y) -> div(x, y)
if_active(true, x, y) -> mark(x)
if_active(false, x, y) -> mark(y)
if_active(x, y, z) -> if(x, y, z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

MINUS_ACTIVE(s(x), s(y)) -> MINUS_ACTIVE(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:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

minus_active(0, y) -> 0
minus_active(s(x), s(y)) -> minus_active(x, y)
minus_active(x, y) -> minus(x, y)
mark(0) -> 0
mark(s(x)) -> s(mark(x))
mark(minus(x, y)) -> minus_active(x, y)
mark(ge(x, y)) -> ge_active(x, y)
mark(div(x, y)) -> div_active(mark(x), y)
mark(if(x, y, z)) -> if_active(mark(x), y, z)
ge_active(x, 0) -> true
ge_active(0, s(y)) -> false
ge_active(s(x), s(y)) -> ge_active(x, y)
ge_active(x, y) -> ge(x, y)
div_active(0, s(y)) -> 0
div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
div_active(x, y) -> div(x, y)
if_active(true, x, y) -> mark(x)
if_active(false, x, y) -> mark(y)
if_active(x, y, z) -> if(x, 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

         ↳Nar

Dependency Pair:

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

Rules:

minus_active(0, y) -> 0
minus_active(s(x), s(y)) -> minus_active(x, y)
minus_active(x, y) -> minus(x, y)
mark(0) -> 0
mark(s(x)) -> s(mark(x))
mark(minus(x, y)) -> minus_active(x, y)
mark(ge(x, y)) -> ge_active(x, y)
mark(div(x, y)) -> div_active(mark(x), y)
mark(if(x, y, z)) -> if_active(mark(x), y, z)
ge_active(x, 0) -> true
ge_active(0, s(y)) -> false
ge_active(s(x), s(y)) -> ge_active(x, y)
ge_active(x, y) -> ge(x, y)
div_active(0, s(y)) -> 0
div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
div_active(x, y) -> div(x, y)
if_active(true, x, y) -> mark(x)
if_active(false, x, y) -> mark(y)
if_active(x, y, z) -> if(x, y, z)

Strategy:

innermost

The following dependency pair can be strictly oriented:

GE_ACTIVE(s(x), s(y)) -> GE_ACTIVE(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:

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

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

Rules:

minus_active(0, y) -> 0
minus_active(s(x), s(y)) -> minus_active(x, y)
minus_active(x, y) -> minus(x, y)
mark(0) -> 0
mark(s(x)) -> s(mark(x))
mark(minus(x, y)) -> minus_active(x, y)
mark(ge(x, y)) -> ge_active(x, y)
mark(div(x, y)) -> div_active(mark(x), y)
mark(if(x, y, z)) -> if_active(mark(x), y, z)
ge_active(x, 0) -> true
ge_active(0, s(y)) -> false
ge_active(s(x), s(y)) -> ge_active(x, y)
ge_active(x, y) -> ge(x, y)
div_active(0, s(y)) -> 0
div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
div_active(x, y) -> div(x, y)
if_active(true, x, y) -> mark(x)
if_active(false, x, y) -> mark(y)
if_active(x, y, z) -> if(x, 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

         ↳Narrowing Transformation

Dependency Pairs:

MARK(if(x, y, z)) -> MARK(x)
IF_ACTIVE(false, x, y) -> MARK(y)
MARK(if(x, y, z)) -> IF_ACTIVE(mark(x), y, z)
MARK(div(x, y)) -> MARK(x)
IF_ACTIVE(true, x, y) -> MARK(x)
DIV_ACTIVE(s(x), s(y)) -> IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
MARK(div(x, y)) -> DIV_ACTIVE(mark(x), y)
MARK(s(x)) -> MARK(x)

Rules:

minus_active(0, y) -> 0
minus_active(s(x), s(y)) -> minus_active(x, y)
minus_active(x, y) -> minus(x, y)
mark(0) -> 0
mark(s(x)) -> s(mark(x))
mark(minus(x, y)) -> minus_active(x, y)
mark(ge(x, y)) -> ge_active(x, y)
mark(div(x, y)) -> div_active(mark(x), y)
mark(if(x, y, z)) -> if_active(mark(x), y, z)
ge_active(x, 0) -> true
ge_active(0, s(y)) -> false
ge_active(s(x), s(y)) -> ge_active(x, y)
ge_active(x, y) -> ge(x, y)
div_active(0, s(y)) -> 0
div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
div_active(x, y) -> div(x, y)
if_active(true, x, y) -> mark(x)
if_active(false, x, y) -> mark(y)
if_active(x, y, z) -> if(x, y, z)

Strategy:

innermost

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

MARK(div(x, y)) -> DIV_ACTIVE(mark(x), y)

six new Dependency Pairs are created:

MARK(div(0, y)) -> DIV_ACTIVE(0, y)
MARK(div(s(x''), y)) -> DIV_ACTIVE(s(mark(x'')), y)
MARK(div(minus(x'', y''), y)) -> DIV_ACTIVE(minus_active(x'', y''), y)
MARK(div(ge(x'', y''), y)) -> DIV_ACTIVE(ge_active(x'', y''), y)
MARK(div(div(x'', y''), y)) -> DIV_ACTIVE(div_active(mark(x''), y''), y)
MARK(div(if(x'', y'', z'), y)) -> DIV_ACTIVE(if_active(mark(x''), y'', z'), y)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

MARK(div(if(x'', y'', z'), y)) -> DIV_ACTIVE(if_active(mark(x''), y'', z'), y)
MARK(div(div(x'', y''), y)) -> DIV_ACTIVE(div_active(mark(x''), y''), y)
MARK(div(ge(x'', y''), y)) -> DIV_ACTIVE(ge_active(x'', y''), y)
MARK(div(minus(x'', y''), y)) -> DIV_ACTIVE(minus_active(x'', y''), y)
IF_ACTIVE(false, x, y) -> MARK(y)
DIV_ACTIVE(s(x), s(y)) -> IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
MARK(div(s(x''), y)) -> DIV_ACTIVE(s(mark(x'')), y)
IF_ACTIVE(true, x, y) -> MARK(x)
MARK(if(x, y, z)) -> IF_ACTIVE(mark(x), y, z)
MARK(div(x, y)) -> MARK(x)
MARK(s(x)) -> MARK(x)
MARK(if(x, y, z)) -> MARK(x)

Rules:

minus_active(0, y) -> 0
minus_active(s(x), s(y)) -> minus_active(x, y)
minus_active(x, y) -> minus(x, y)
mark(0) -> 0
mark(s(x)) -> s(mark(x))
mark(minus(x, y)) -> minus_active(x, y)
mark(ge(x, y)) -> ge_active(x, y)
mark(div(x, y)) -> div_active(mark(x), y)
mark(if(x, y, z)) -> if_active(mark(x), y, z)
ge_active(x, 0) -> true
ge_active(0, s(y)) -> false
ge_active(s(x), s(y)) -> ge_active(x, y)
ge_active(x, y) -> ge(x, y)
div_active(0, s(y)) -> 0
div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
div_active(x, y) -> div(x, y)
if_active(true, x, y) -> mark(x)
if_active(false, x, y) -> mark(y)
if_active(x, y, z) -> if(x, y, z)

Strategy:

innermost

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

MARK(if(x, y, z)) -> IF_ACTIVE(mark(x), y, z)

six new Dependency Pairs are created:

MARK(if(0, y, z)) -> IF_ACTIVE(0, y, z)
MARK(if(s(x''), y, z)) -> IF_ACTIVE(s(mark(x'')), y, z)
MARK(if(minus(x'', y''), y, z)) -> IF_ACTIVE(minus_active(x'', y''), y, z)
MARK(if(ge(x'', y''), y, z)) -> IF_ACTIVE(ge_active(x'', y''), y, z)
MARK(if(div(x'', y''), y, z)) -> IF_ACTIVE(div_active(mark(x''), y''), y, z)
MARK(if(if(x'', y'', z''), y, z)) -> IF_ACTIVE(if_active(mark(x''), y'', z''), y, 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 6

             ↳Nar

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

MARK(if(if(x'', y'', z''), y, z)) -> IF_ACTIVE(if_active(mark(x''), y'', z''), y, z)
MARK(if(div(x'', y''), y, z)) -> IF_ACTIVE(div_active(mark(x''), y''), y, z)
MARK(if(ge(x'', y''), y, z)) -> IF_ACTIVE(ge_active(x'', y''), y, z)
IF_ACTIVE(false, x, y) -> MARK(y)
MARK(if(minus(x'', y''), y, z)) -> IF_ACTIVE(minus_active(x'', y''), y, z)
MARK(div(div(x'', y''), y)) -> DIV_ACTIVE(div_active(mark(x''), y''), y)
MARK(div(ge(x'', y''), y)) -> DIV_ACTIVE(ge_active(x'', y''), y)
MARK(div(minus(x'', y''), y)) -> DIV_ACTIVE(minus_active(x'', y''), y)
MARK(div(s(x''), y)) -> DIV_ACTIVE(s(mark(x'')), y)
MARK(if(x, y, z)) -> MARK(x)
MARK(div(x, y)) -> MARK(x)
MARK(s(x)) -> MARK(x)
IF_ACTIVE(true, x, y) -> MARK(x)
DIV_ACTIVE(s(x), s(y)) -> IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
MARK(div(if(x'', y'', z'), y)) -> DIV_ACTIVE(if_active(mark(x''), y'', z'), y)

Rules:

minus_active(0, y) -> 0
minus_active(s(x), s(y)) -> minus_active(x, y)
minus_active(x, y) -> minus(x, y)
mark(0) -> 0
mark(s(x)) -> s(mark(x))
mark(minus(x, y)) -> minus_active(x, y)
mark(ge(x, y)) -> ge_active(x, y)
mark(div(x, y)) -> div_active(mark(x), y)
mark(if(x, y, z)) -> if_active(mark(x), y, z)
ge_active(x, 0) -> true
ge_active(0, s(y)) -> false
ge_active(s(x), s(y)) -> ge_active(x, y)
ge_active(x, y) -> ge(x, y)
div_active(0, s(y)) -> 0
div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
div_active(x, y) -> div(x, y)
if_active(true, x, y) -> mark(x)
if_active(false, x, y) -> mark(y)
if_active(x, y, z) -> if(x, y, z)

Strategy:

innermost

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

DIV_ACTIVE(s(x), s(y)) -> IF_ACTIVE(ge_active(x, y), s(div(minus(x, y), s(y))), 0)

four new Dependency Pairs are created:

DIV_ACTIVE(s(x''), s(0)) -> IF_ACTIVE(true, s(div(minus(x'', 0), s(0))), 0)
DIV_ACTIVE(s(0), s(s(y''))) -> IF_ACTIVE(false, s(div(minus(0, s(y'')), s(s(y'')))), 0)
DIV_ACTIVE(s(s(x'')), s(s(y''))) -> IF_ACTIVE(ge_active(x'', y''), s(div(minus(s(x''), s(y'')), s(s(y'')))), 0)
DIV_ACTIVE(s(x''), s(y'')) -> IF_ACTIVE(ge(x'', y''), s(div(minus(x'', y''), s(y''))), 0)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

MARK(if(div(x'', y''), y, z)) -> IF_ACTIVE(div_active(mark(x''), y''), y, z)
MARK(if(ge(x'', y''), y, z)) -> IF_ACTIVE(ge_active(x'', y''), y, z)
MARK(if(minus(x'', y''), y, z)) -> IF_ACTIVE(minus_active(x'', y''), y, z)
MARK(div(if(x'', y'', z'), y)) -> DIV_ACTIVE(if_active(mark(x''), y'', z'), y)
MARK(div(div(x'', y''), y)) -> DIV_ACTIVE(div_active(mark(x''), y''), y)
MARK(div(ge(x'', y''), y)) -> DIV_ACTIVE(ge_active(x'', y''), y)
DIV_ACTIVE(s(s(x'')), s(s(y''))) -> IF_ACTIVE(ge_active(x'', y''), s(div(minus(s(x''), s(y'')), s(s(y'')))), 0)
MARK(div(minus(x'', y''), y)) -> DIV_ACTIVE(minus_active(x'', y''), y)
IF_ACTIVE(false, x, y) -> MARK(y)
DIV_ACTIVE(s(0), s(s(y''))) -> IF_ACTIVE(false, s(div(minus(0, s(y'')), s(s(y'')))), 0)
DIV_ACTIVE(s(x''), s(0)) -> IF_ACTIVE(true, s(div(minus(x'', 0), s(0))), 0)
MARK(div(s(x''), y)) -> DIV_ACTIVE(s(mark(x'')), y)
MARK(if(x, y, z)) -> MARK(x)
MARK(div(x, y)) -> MARK(x)
MARK(s(x)) -> MARK(x)
IF_ACTIVE(true, x, y) -> MARK(x)
MARK(if(if(x'', y'', z''), y, z)) -> IF_ACTIVE(if_active(mark(x''), y'', z''), y, z)

Rules:

minus_active(0, y) -> 0
minus_active(s(x), s(y)) -> minus_active(x, y)
minus_active(x, y) -> minus(x, y)
mark(0) -> 0
mark(s(x)) -> s(mark(x))
mark(minus(x, y)) -> minus_active(x, y)
mark(ge(x, y)) -> ge_active(x, y)
mark(div(x, y)) -> div_active(mark(x), y)
mark(if(x, y, z)) -> if_active(mark(x), y, z)
ge_active(x, 0) -> true
ge_active(0, s(y)) -> false
ge_active(s(x), s(y)) -> ge_active(x, y)
ge_active(x, y) -> ge(x, y)
div_active(0, s(y)) -> 0
div_active(s(x), s(y)) -> if_active(ge_active(x, y), s(div(minus(x, y), s(y))), 0)
div_active(x, y) -> div(x, y)
if_active(true, x, y) -> mark(x)
if_active(false, x, y) -> mark(y)
if_active(x, y, z) -> if(x, y, z)

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:41 minutes