Termination Proof using AProVE

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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))
QUOT(x, s(y)) -> LE(s(y), x)
IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
IF_QUOT(true, x, 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:

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

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 8 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →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

         ↳Usable Rules (Innermost)

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 8 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Size-Change Principle

       →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

         ↳UsableRules

       →DP Problem 3

         ↳Usable Rules (Innermost)

Dependency Pairs:

IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

Rules:

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

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Narrowing Transformation

Dependency Pairs:

IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

Rules:

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

QUOT(x, s(y)) -> IF_QUOT(le(s(y), x), x, s(y))

two new Dependency Pairs are created:

QUOT(s(y''), s(y0)) -> IF_QUOT(le(y0, y''), s(y''), s(y0))
QUOT(0, s(y')) -> IF_QUOT(false, 0, s(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 7

                 ↳Narrowing Transformation

Dependency Pairs:

QUOT(s(y''), s(y0)) -> IF_QUOT(le(y0, y''), s(y''), s(y0))
IF_QUOT(true, x, y) -> QUOT(minus(x, y), y)

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
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_QUOT(true, x, y) -> QUOT(minus(x, y), y)

two new Dependency Pairs are created:

IF_QUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))
IF_QUOT(true, x'', 0) -> QUOT(x'', 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 8

                 ↳Negative Polynomial Order

Dependency Pairs:

IF_QUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))
QUOT(s(y''), s(y0)) -> IF_QUOT(le(y0, y''), s(y''), s(y0))

Rules:

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

Strategy:

innermost

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

IF_QUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), 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
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_QUOT(x₁, ..., x₃) ) = x₂

POL( s(x₁) ) = x₁ + 1

POL( QUOT(x₁, x₂) ) = x₁

POL( minus(x₁, x₂) ) = 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

             ↳Nar

...

               →DP Problem 9

                 ↳Dependency Graph

Dependency Pair:

QUOT(s(y''), s(y0)) -> IF_QUOT(le(y0, y''), s(y''), s(y0))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> 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:
0:06 minutes