Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

-'(s(x), s(y)) -> -'(x, y)
<='(s(x), s(y)) -> <='(x, y)
PERFECTP(s(x)) -> F(x, s(0), s(x), s(x))
F(s(x), 0, z, u) -> F(x, u, -(z, s(x)), u)
F(s(x), 0, z, u) -> -'(z, s(x))
F(s(x), s(y), z, u) -> IF(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))
F(s(x), s(y), z, u) -> <='(x, y)
F(s(x), s(y), z, u) -> F(s(x), -(y, x), z, u)
F(s(x), s(y), z, u) -> -'(y, x)
F(s(x), s(y), z, u) -> F(x, u, z, u)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

As we are in the innermost case, we can delete all 13 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

<='(s(x), s(y)) -> <='(x, y)

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), f(x, u, z, u))

Strategy:

innermost

As we are in the innermost case, we can delete all 13 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

<='(s(x), s(y)) -> <='(x, y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:

<='(s(x), s(y)) -> <='(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:

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

Rules:

-(x, 0) -> x
-(s(x), s(y)) -> -(x, y)
<=(0, y) -> true
<=(s(x), 0) -> false
<=(s(x), s(y)) -> <=(x, y)
if(true, x, y) -> x
if(false, x, y) -> y
perfectp(0) -> false
perfectp(s(x)) -> f(x, s(0), s(x), s(x))
f(0, y, 0, u) -> true
f(0, y, s(z), u) -> false
f(s(x), 0, z, u) -> f(x, u, -(z, s(x)), u)
f(s(x), s(y), z, u) -> if(<=(x, y), f(s(x), -(y, x), z, u), 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 6

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

Rules:

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

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

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

Used ordering:
Polynomial Order with Interpretation:

POL( F(x₁, ..., x₄) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( -(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 6

             ↳Neg POLO

...

               →DP Problem 7

                 ↳Negative Polynomial Order

Dependency Pair:

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

Rules:

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

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

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

Used ordering:
Polynomial Order with Interpretation:

POL( F(x₁, ..., x₄) ) = x₂

POL( s(x₁) ) = x₁ + 1

POL( -(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 6

             ↳Neg POLO

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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