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

         ↳Argument Filtering and Ordering

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

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

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

MINUS(x₁, x₂) -> MINUS(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

           →DP Problem 4

             ↳Dependency Graph

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳Argument Filtering and Ordering

       →DP Problem 3

         ↳Nar

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

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:

POL(LE(x₁, x₂)) = x₁ + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.
Used Argument Filtering System:

LE(x₁, x₂) -> LE(x₁, x₂)
s(x₁) -> s(x₁)

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

           →DP Problem 5

             ↳Dependency Graph

       →DP Problem 3

         ↳Nar

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳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(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

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳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(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

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 7

                 ↳Argument Filtering and Ordering

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(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

The following dependency pair can be strictly oriented:

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

The following usable rules for innermost w.r.t. to the AFS can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(QUOT(x₁, x₂)) = 1 + x₁ + x₂

POL(false) = 0

POL(true) = 0

POL(s(x₁)) = 1 + x₁

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

POL(le) = 0

resulting in one new DP problem.
Used Argument Filtering System:

QUOT(x₁, x₂) -> QUOT(x₁, x₂)
IF_QUOT(x₁, x₂, x₃) -> IF_QUOT(x₁, x₂, x₃)
s(x₁) -> s(x₁)
le(x₁, x₂) -> le
minus(x₁, x₂) -> x₁

     ↳DPs

       →DP Problem 1

         ↳AFS

       →DP Problem 2

         ↳AFS

       →DP Problem 3

         ↳Nar

           →DP Problem 6

             ↳Nar

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

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

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes

POL(MINUS(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(LE(x₁, x₂))	= x₁ + x₂
POL(s(x₁))	= 1 + x₁

POL(QUOT(x₁, x₂))	= 1 + x₁ + x₂
POL(false)	= 0
POL(true)	= 0
POL(s(x₁))	= 1 + x₁
POL(IF_QUOT(x₁, x₂, x₃))	= x₁ + x₂ + x₃
POL(le)	= 0