Termination Proof using AProVE

Term Rewriting System R:
[y, x]
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) -> gcd(minus(x, y), y)
if_gcd(false, x, y) -> gcd(minus(y, x), x)

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

LE(s(x), s(y)) -> LE(x, y)
MINUS(s(x), s(y)) -> MINUS(x, y)
GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), s(y))
GCD(s(x), s(y)) -> LE(y, x)
IF_GCD(true, x, y) -> GCD(minus(x, y), y)
IF_GCD(true, x, y) -> MINUS(x, y)
IF_GCD(false, x, y) -> GCD(minus(y, x), x)
IF_GCD(false, x, y) -> MINUS(y, x)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) -> gcd(minus(x, y), y)
if_gcd(false, x, y) -> gcd(minus(y, x), x)

Strategy:

innermost

As we are in the innermost case, we can delete all 11 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →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

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) -> gcd(minus(x, y), y)
if_gcd(false, x, y) -> gcd(minus(y, x), x)

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 5

             ↳Size-Change Principle

       →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

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

Dependency Pairs:

IF_GCD(false, x, y) -> GCD(minus(y, x), x)
IF_GCD(true, x, y) -> GCD(minus(x, y), y)
GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), s(y))

Rules:

le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
minus(s(x), s(y)) -> minus(x, y)
gcd(0, y) -> y
gcd(s(x), 0) -> s(x)
gcd(s(x), s(y)) -> if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) -> gcd(minus(x, y), y)
if_gcd(false, x, y) -> gcd(minus(y, x), x)

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

IF_GCD(false, x, y) -> GCD(minus(y, x), x)
IF_GCD(true, x, y) -> GCD(minus(x, y), y)
GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), s(y))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF_GCD(false, x, y) -> GCD(minus(y, x), x)

three new Dependency Pairs are created:

IF_GCD(false, s(y''), s(x'')) -> GCD(minus(x'', y''), s(y''))
IF_GCD(false, 0, y') -> GCD(y', 0)
IF_GCD(false, x'', 0) -> GCD(0, x'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Narrowing Transformation

Dependency Pairs:

IF_GCD(false, s(y''), s(x'')) -> GCD(minus(x'', y''), s(y''))
GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), s(y))
IF_GCD(true, x, y) -> GCD(minus(x, y), y)

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IF_GCD(true, x, y) -> GCD(minus(x, y), y)

three new Dependency Pairs are created:

IF_GCD(true, s(x''), s(y'')) -> GCD(minus(x'', y''), s(y''))
IF_GCD(true, x'', 0) -> GCD(x'', 0)
IF_GCD(true, 0, y') -> GCD(0, y')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Narrowing Transformation

Dependency Pairs:

IF_GCD(true, s(x''), s(y'')) -> GCD(minus(x'', y''), s(y''))
GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), s(y))
IF_GCD(false, s(y''), s(x'')) -> GCD(minus(x'', y''), s(y''))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

GCD(s(x), s(y)) -> IF_GCD(le(y, x), s(x), s(y))

three new Dependency Pairs are created:

GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))
GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))
GCD(s(0), s(s(x''))) -> IF_GCD(false, s(0), s(s(x'')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 9

                 ↳Instantiation Transformation

Dependency Pairs:

GCD(s(0), s(s(x''))) -> IF_GCD(false, s(0), s(s(x'')))
GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))
IF_GCD(false, s(y''), s(x'')) -> GCD(minus(x'', y''), s(y''))
GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))
IF_GCD(true, s(x''), s(y'')) -> GCD(minus(x'', y''), s(y''))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_GCD(false, s(y''), s(x'')) -> GCD(minus(x'', y''), s(y''))

two new Dependency Pairs are created:

IF_GCD(false, s(s(y'''')), s(s(x''''))) -> GCD(minus(s(x''''), s(y'''')), s(s(y'''')))
IF_GCD(false, s(0), s(s(x''''))) -> GCD(minus(s(x''''), 0), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 10

                 ↳Rewriting Transformation

Dependency Pairs:

IF_GCD(false, s(s(y'''')), s(s(x''''))) -> GCD(minus(s(x''''), s(y'''')), s(s(y'''')))
GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))
IF_GCD(true, s(x''), s(y'')) -> GCD(minus(x'', y''), s(y''))
GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))
IF_GCD(false, s(0), s(s(x''''))) -> GCD(minus(s(x''''), 0), s(0))
GCD(s(0), s(s(x''))) -> IF_GCD(false, s(0), s(s(x'')))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

IF_GCD(false, s(0), s(s(x''''))) -> GCD(minus(s(x''''), 0), s(0))

one new Dependency Pair is created:

IF_GCD(false, s(0), s(s(x''''))) -> GCD(s(x''''), s(0))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 11

                 ↳Instantiation Transformation

Dependency Pairs:

IF_GCD(false, s(0), s(s(x''''))) -> GCD(s(x''''), s(0))
GCD(s(0), s(s(x''))) -> IF_GCD(false, s(0), s(s(x'')))
GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))
IF_GCD(true, s(x''), s(y'')) -> GCD(minus(x'', y''), s(y''))
GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))
IF_GCD(false, s(s(y'''')), s(s(x''''))) -> GCD(minus(s(x''''), s(y'''')), s(s(y'''')))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

IF_GCD(true, s(x''), s(y'')) -> GCD(minus(x'', y''), s(y''))

two new Dependency Pairs are created:

IF_GCD(true, s(s(y'''')), s(s(x''''))) -> GCD(minus(s(y''''), s(x'''')), s(s(x'''')))
IF_GCD(true, s(x''''), s(0)) -> GCD(minus(x'''', 0), s(0))

The transformation is resulting in two new DP problems:

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 12

                 ↳Usable Rules (Innermost)

Dependency Pairs:

IF_GCD(true, s(x''''), s(0)) -> GCD(minus(x'''', 0), s(0))
GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

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 6

             ↳Nar

...

               →DP Problem 14

                 ↳Modular Removal of Rules

Dependency Pairs:

IF_GCD(true, s(x''''), s(0)) -> GCD(minus(x'''', 0), s(0))
GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))

Rules:

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

Strategy:

innermost

We have the following set of usable rules:

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

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(0) = 0

POL(GCD(x₁, x₂)) = x₁ + x₂

POL(minus(x₁, x₂)) = x₁ + x₂

POL(IF_GCD(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(true) = 0

POL(s(x₁)) = 1 + x₁

We have the following set D of usable symbols: {0, GCD, minus, IF_GCD, true, s}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

IF_GCD(true, s(x''''), s(0)) -> GCD(minus(x'''', 0), s(0))

No Rules can be deleted.

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 15

                 ↳Modular Removal of Rules

Dependency Pair:

GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))

Rules:

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

Strategy:

innermost

We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(0) = 0

POL(GCD(x₁, x₂)) = x₁ + x₂

POL(IF_GCD(x₁, x₂, x₃)) = x₁ + x₂ + x₃

POL(true) = 0

POL(s(x₁)) = x₁

We have the following set D of usable symbols: {0, IF_GCD, true, s}
The following Dependency Pairs can be deleted as they contain symbols in their lhs which do not occur in D:

GCD(s(x'), s(0)) -> IF_GCD(true, s(x'), s(0))

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 13

                 ↳Negative Polynomial Order

Dependency Pairs:

IF_GCD(true, s(s(y'''')), s(s(x''''))) -> GCD(minus(s(y''''), s(x'''')), s(s(x'''')))
IF_GCD(false, s(s(y'''')), s(s(x''''))) -> GCD(minus(s(x''''), s(y'''')), s(s(y'''')))
GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

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

IF_GCD(true, s(s(y'''')), s(s(x''''))) -> GCD(minus(s(y''''), s(x'''')), s(s(x'''')))
IF_GCD(false, s(s(y'''')), s(s(x''''))) -> GCD(minus(s(x''''), s(y'''')), s(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
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Used ordering:
Polynomial Order with Interpretation:

POL( IF_GCD(x₁, ..., x₃) ) = x₂ + x₃

POL( s(x₁) ) = x₁ + 1

POL( GCD(x₁, x₂) ) = x₁ + x₂

POL( minus(x₁, x₂) ) = x₁

POL( 0 ) = 0

POL( le(x₁, x₂) ) = 0

POL( true ) = 0

POL( false ) = 0

This results in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Nar

...

               →DP Problem 16

                 ↳Dependency Graph

Dependency Pair:

GCD(s(s(y'')), s(s(x''))) -> IF_GCD(le(x'', y''), s(s(y'')), s(s(x'')))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
minus(0, x) -> 0
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:07 minutes

POL(0)	= 0
POL(GCD(x₁, x₂))	= x₁ + x₂
POL(minus(x₁, x₂))	= x₁ + x₂
POL(IF_GCD(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(true)	= 0
POL(s(x₁))	= 1 + x₁