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

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

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

Strategy:

innermost

As we are in the innermost case, we can delete all 8 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(double, app(s, x)) -> APP(double, x)

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 8 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 6

             ↳A-Transformation

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

APP(double, app(s, x)) -> APP(double, x)

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 2

         ↳UsableRules

           →DP Problem 6

             ↳ATrans

...

               →DP Problem 7

                 ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

DOUBLE(s(x)) -> DOUBLE(x)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

DOUBLE(s(x)) -> DOUBLE(x)

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

{1}	,	{1}
1	>	1

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

{1}	,	{1}
1	>	1

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

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

Dependency Pairs:

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

Rules:

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

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 3

         ↳UsableRules

           →DP Problem 8

             ↳A-Transformation

Dependency Pairs:

Rules:

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

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 8

             ↳ATrans

...

               →DP Problem 9

                 ↳Argument Filtering and Ordering

Dependency Pairs:

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

Rules:

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

Strategy:

innermost

The following dependency pairs can be strictly oriented:

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

The following usable rules w.r.t. the AFS can be oriented:

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

Used ordering: Lexicographic Path Order with Precedence:

trivial

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

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

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 8

             ↳ATrans

...

               →DP Problem 10

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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