Termination Proof using AProVE

Term Rewriting System R:
[y, x, u, z]
app(app(minus, 0), y) -> 0
app(app(minus, app(s, x)), 0) -> app(s, x)
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(le, 0), y) -> true
app(app(le, app(s, x)), 0) -> false
app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(perfectp, 0) -> false
app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) -> true
app(app(app(app(f, 0), y), app(s, z)), u) -> false
app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))

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(le, app(s, x)), app(s, y)) -> APP(app(le, x), y)
APP(app(le, app(s, x)), app(s, y)) -> APP(le, x)
APP(perfectp, app(s, x)) -> APP(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
APP(perfectp, app(s, x)) -> APP(app(app(f, x), app(s, 0)), app(s, x))
APP(perfectp, app(s, x)) -> APP(app(f, x), app(s, 0))
APP(perfectp, app(s, x)) -> APP(f, x)
APP(perfectp, app(s, x)) -> APP(s, 0)
APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(app(f, x), u), app(app(minus, z), app(s, x)))
APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(f, x), u)
APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(f, x)
APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(minus, z), app(s, x))
APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(minus, z)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(if, app(app(le, x), y))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(le, x), y)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(le, x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(f, app(s, x)), app(app(minus, y), x)), z)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(f, app(s, x)), app(app(minus, y), x))
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(minus, y), x)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(minus, y)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, x), u), z), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(f, x), u), z)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(f, x), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(f, x)

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, 0), y) -> 0
app(app(minus, app(s, x)), 0) -> app(s, x)
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(le, 0), y) -> true
app(app(le, app(s, x)), 0) -> false
app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(perfectp, 0) -> false
app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) -> true
app(app(app(app(f, 0), y), app(s, z)), u) -> false
app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))

Strategy:

innermost

As we are in the innermost case, we can delete all 14 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(le, app(s, x)), app(s, y)) -> APP(app(le, x), y)

Rules:

app(app(minus, 0), y) -> 0
app(app(minus, app(s, x)), 0) -> app(s, x)
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(le, 0), y) -> true
app(app(le, app(s, x)), 0) -> false
app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(perfectp, 0) -> false
app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) -> true
app(app(app(app(f, 0), y), app(s, z)), u) -> false
app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))

Strategy:

innermost

As we are in the innermost case, we can delete all 14 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(le, app(s, x)), app(s, y)) -> APP(app(le, 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 2

         ↳UsableRules

           →DP Problem 6

             ↳ATrans

...

               →DP Problem 7

                 ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

Dependency Pairs:

APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, x), u), z), u)
APP(app(app(app(f, app(s, x)), 0), z), u) -> APP(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
APP(app(app(app(f, app(s, x)), app(s, y)), z), u) -> APP(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)

Rules:

app(app(minus, 0), y) -> 0
app(app(minus, app(s, x)), 0) -> app(s, x)
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(le, 0), y) -> true
app(app(le, app(s, x)), 0) -> false
app(app(le, app(s, x)), app(s, y)) -> app(app(le, x), y)
app(app(app(if, true), x), y) -> x
app(app(app(if, false), x), y) -> y
app(perfectp, 0) -> false
app(perfectp, app(s, x)) -> app(app(app(app(f, x), app(s, 0)), app(s, x)), app(s, x))
app(app(app(app(f, 0), y), 0), u) -> true
app(app(app(app(f, 0), y), app(s, z)), u) -> false
app(app(app(app(f, app(s, x)), 0), z), u) -> app(app(app(app(f, x), u), app(app(minus, z), app(s, x))), u)
app(app(app(app(f, app(s, x)), app(s, y)), z), u) -> app(app(app(if, app(app(le, x), y)), app(app(app(app(f, app(s, x)), app(app(minus, y), x)), z), u)), app(app(app(app(f, x), u), z), u))

Strategy:

innermost

As we are in the innermost case, we can delete all 11 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, 0), y) -> 0
app(app(minus, app(s, x)), app(s, y)) -> app(app(minus, x), y)
app(app(minus, app(s, x)), 0) -> app(s, 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

                 ↳Negative Polynomial Order

Dependency Pairs:

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)
F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)

Rules:

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

Strategy:

innermost

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

F(s(x), s(y), z, u) -> F(x, u, z, u)
F(s(x), 0, z, u) -> F(x, u, minus(z, s(x)), u)

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

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

Used ordering:
Polynomial Order with Interpretation:

POL( F(x₁, ..., x₄) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( minus(x₁, x₂) ) = x₁

POL( 0 ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 8

             ↳ATrans

...

               →DP Problem 10

                 ↳Negative Polynomial Order

Dependency Pair:

F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)

Rules:

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

Strategy:

innermost

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

F(s(x), s(y), z, u) -> F(s(x), minus(y, x), z, u)

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

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

Used ordering:
Polynomial Order with Interpretation:

POL( F(x₁, ..., x₄) ) = x₂

POL( s(x₁) ) = x₁ + 1

POL( minus(x₁, x₂) ) = x₁

POL( 0 ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 8

             ↳ATrans

...

               →DP Problem 11

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:02 minutes