Termination Proof using AProVE

Term Rewriting System R:
[x, y]
app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(log, app(s, 0)) -> 0
app(log, app(s, app(s, x))) -> app(s, app(log, app(s, app(app(quot, x), app(s, app(s, 0))))))

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(minus, app(s, x)), app(s, y)) -> APP(minus, x)
APP(app(quot, app(s, x)), app(s, y)) -> APP(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(quot, app(app(minus, x), y))
APP(app(quot, app(s, x)), app(s, y)) -> APP(app(minus, x), y)
APP(app(quot, app(s, x)), app(s, y)) -> APP(minus, x)
APP(log, app(s, app(s, x))) -> APP(s, app(log, app(s, app(app(quot, x), app(s, app(s, 0))))))
APP(log, app(s, app(s, x))) -> APP(log, app(s, app(app(quot, x), app(s, app(s, 0)))))
APP(log, app(s, app(s, x))) -> APP(s, app(app(quot, x), app(s, app(s, 0))))
APP(log, app(s, app(s, x))) -> APP(app(quot, x), app(s, app(s, 0)))
APP(log, app(s, app(s, x))) -> APP(quot, x)
APP(log, app(s, app(s, x))) -> APP(s, app(s, 0))
APP(log, app(s, app(s, x))) -> APP(s, 0)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)

Rules:

app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(log, app(s, 0)) -> 0
app(log, app(s, app(s, x))) -> app(s, app(log, app(s, app(app(quot, x), app(s, app(s, 0))))))

Strategy:

innermost

As we are in the innermost case, we can delete all 6 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳A-Transformation

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

APP(app(minus, app(s, x)), app(s, y)) -> APP(app(minus, x), y)

Rule:

none

Strategy:

innermost

We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳ATrans

...

               →DP Problem 5

                 ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

and get the following Size-Change Graph(s):

{1}	,	{1}
1	>	1
2	>	2

which lead(s) to this/these maximal multigraph(s):

{1}	,	{1}
1	>	1
2	>	2

D_P: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.

trivial

with Argument Filtering System:

s(x₁) -> s(x₁)

We obtain no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))

Rules:

app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(log, app(s, 0)) -> 0
app(log, app(s, app(s, x))) -> app(s, app(log, app(s, app(app(quot, x), app(s, app(s, 0))))))

Strategy:

innermost

As we are in the innermost case, we can delete all 4 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 6

             ↳A-Transformation

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

APP(app(quot, app(s, x)), app(s, y)) -> APP(app(quot, app(app(minus, x), y)), app(s, y))

Rules:

app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)

Strategy:

innermost

We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 6

             ↳ATrans

...

               →DP Problem 7

                 ↳Negative Polynomial Order

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

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

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x

Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x₁, x₂) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( minus(x₁, x₂) ) = x₁

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 6

             ↳ATrans

...

               →DP Problem 8

                 ↳Dependency Graph

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

Dependency Pair:

APP(log, app(s, app(s, x))) -> APP(log, app(s, app(app(quot, x), app(s, app(s, 0)))))

Rules:

app(app(minus, x), 0) -> x
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(quot, 0), app(s, y)) -> 0
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(log, app(s, 0)) -> 0
app(log, app(s, app(s, x))) -> app(s, app(log, app(s, app(app(quot, x), app(s, app(s, 0))))))

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 9

             ↳A-Transformation

Dependency Pair:

APP(log, app(s, app(s, x))) -> APP(log, app(s, app(app(quot, x), app(s, app(s, 0)))))

Rules:

app(app(minus, x), 0) -> x
app(app(quot, app(s, x)), app(s, y)) -> app(s, app(app(quot, app(app(minus, x), y)), app(s, y)))
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(quot, 0), app(s, y)) -> 0

Strategy:

innermost

We have an applicative DP problem with proper arity. Thus we can use the A-Transformation to obtain one new DP problem which consists of the A-transformed TRSs.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 9

             ↳ATrans

...

               →DP Problem 10

                 ↳Negative Polynomial Order

Dependency Pair:

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
quot(0, s(y)) -> 0

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

LOG(s(s(x))) -> LOG(s(quot(x, s(s(0)))))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

quot(0, s(y)) -> 0
minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))

Used ordering:
Polynomial Order with Interpretation:

POL( LOG(x₁) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( quot(x₁, x₂) ) = x₁

POL( 0 ) = 0

POL( minus(x₁, x₂) ) = x₁

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 9

             ↳ATrans

...

               →DP Problem 11

                 ↳Dependency Graph

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
quot(0, s(y)) -> 0

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes