Termination Proof using AProVE

Term Rewriting System R:
[x, y]
p(0) -> 0
p(s(x)) -> x
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, 0) -> x
minus(x, s(y)) -> if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) -> x
if(false, x, y) -> 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(x, s(y)) -> IF(le(x, s(y)), 0, p(minus(x, p(s(y)))))
MINUS(x, s(y)) -> LE(x, s(y))
MINUS(x, s(y)) -> P(minus(x, p(s(y))))
MINUS(x, s(y)) -> MINUS(x, p(s(y)))
MINUS(x, s(y)) -> P(s(y))

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Rw

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳Rw

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 4

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳Rw

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(LE(x₁, x₂)) = 1 + x₁

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 3

             ↳FwdInst

...

               →DP Problem 5

                 ↳Dependency Graph

       →DP Problem 2

         ↳Rw

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rewriting Transformation

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳Forward Instantiation Transformation

Dependency Pair:

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

Rules:

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

Strategy:

innermost

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

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

one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 7

                 ↳Polynomial Ordering

Dependency Pair:

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

Rules:

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(MINUS(x₁, x₂)) = 1 + x₂

POL(s(x₁)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Rw

           →DP Problem 6

             ↳FwdInst

...

               →DP Problem 8

                 ↳Dependency Graph

Dependency Pair:

Rules:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

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