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(f, 0) -> app(s, 0)
app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) -> 0
app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))

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(f, 0) -> APP(s, 0)
APP(f, app(s, x)) -> APP(app(minus, app(s, x)), app(g, app(f, x)))
APP(f, app(s, x)) -> APP(minus, app(s, x))
APP(f, app(s, x)) -> APP(g, app(f, x))
APP(f, app(s, x)) -> APP(f, x)
APP(g, app(s, x)) -> APP(app(minus, app(s, x)), app(f, app(g, x)))
APP(g, app(s, x)) -> APP(minus, app(s, x))
APP(g, app(s, x)) -> APP(f, app(g, x))
APP(g, app(s, x)) -> APP(g, x)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳ATrans

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(f, 0) -> app(s, 0)
app(f, app(s, x)) -> app(app(minus, app(s, x)), app(g, app(f, x)))
app(g, 0) -> 0
app(g, app(s, x)) -> app(app(minus, app(s, x)), app(f, app(g, x)))

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 3

             ↳A-Transformation

       →DP Problem 2

         ↳ATrans

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 3

             ↳ATrans

...

               →DP Problem 4

                 ↳Size-Change Principle

       →DP Problem 2

         ↳ATrans

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

         ↳A-Transformation

Dependency Pairs:

APP(g, app(s, x)) -> APP(g, x)
APP(f, app(s, x)) -> APP(f, x)
APP(g, app(s, x)) -> APP(f, app(g, x))
APP(f, app(s, x)) -> APP(g, app(f, x))

Rules:

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

         ↳ATrans

           →DP Problem 5

             ↳Negative Polynomial Order

Dependency Pairs:

G(s(x)) -> G(x)
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))
F(s(x)) -> G(f(x))

Rules:

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

Strategy:

innermost

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

G(s(x)) -> G(x)
F(s(x)) -> F(x)
G(s(x)) -> F(g(x))

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

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

Used ordering:
Polynomial Order with Interpretation:

POL( G(x₁) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( F(x₁) ) = x₁

POL( f(x₁) ) = x₁ + 1

POL( g(x₁) ) = x₁

POL( 0 ) = 0

POL( minus(x₁, x₂) ) = x₁

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳ATrans

           →DP Problem 5

             ↳Neg POLO

...

               →DP Problem 6

                 ↳Dependency Graph

Dependency Pair:

F(s(x)) -> G(f(x))

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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