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)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(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(x, s(y)) -> PRED(minus(x, y))
MINUS(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, s(x), s(y)) -> GCD(minus(x, y), s(y))
IF_GCD(true, s(x), s(y)) -> MINUS(x, y)
IF_GCD(false, s(x), s(y)) -> GCD(minus(y, x), s(x))
IF_GCD(false, s(x), s(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)
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(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, s(x), s(y)) -> gcd(minus(x, y), s(y))
if_gcd(false, s(x), s(y)) -> gcd(minus(y, x), s(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(x, s(y)) -> MINUS(x, y)

Rules:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

Rules:

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

             ↳Negative Polynomial Order

Dependency Pairs:

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

Rules:

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

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

minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x
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( pred(x₁) ) = x₁

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

             ↳Neg POLO

...

               →DP Problem 7

                 ↳Dependency Graph

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x
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:
0:00 minutes