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(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

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), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
MINUS(s(X), Y) -> LE(s(X), Y)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MINUS(X, Y)

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(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

As we are in the innermost case, we can delete all 9 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 Pairs:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

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

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

Dependency Pairs:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)

Rules:

le(s(X), 0) -> false
le(0, Y) -> true
le(s(X), s(Y)) -> le(X, Y)

Strategy:

innermost

We number the DPs as follows:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)

and get the following Size-Change Graph(s):

{1}	,	{1}
2	>	1
3	=	2

{2}	,	{2}
1	=	2
2	=	3

which lead(s) to this/these maximal multigraph(s):

{2}	,	{1}
1	>	1
2	=	2

{1}	,	{2}
2	>	2
3	=	3

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 Pair:

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Negative Polynomial Order

Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

Rules:

minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
minus(0, Y) -> 0
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(s(X), 0) -> false
le(0, Y) -> true
le(s(X), s(Y)) -> le(X, Y)

Strategy:

innermost

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

QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))

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

minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
minus(0, Y) -> 0
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(s(X), 0) -> false
le(0, Y) -> true
le(s(X), s(Y)) -> le(X, Y)

Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x₁, x₂) ) = x₁

POL( s(x₁) ) = x₁ + 1

POL( minus(x₁, x₂) ) = x₁

POL( ifMinus(x₁, ..., x₃) ) = x₂

POL( 0 ) = 0

POL( le(x₁, x₂) ) = 0

POL( false ) = 0

POL( true ) = 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:

Rules:

minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
minus(0, Y) -> 0
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(s(X), 0) -> false
le(0, Y) -> true
le(s(X), s(Y)) -> le(X, Y)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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