Termination Proof using AProVE

Term Rewriting System R:
[X, Y]
fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

FACT(X) -> IF(zero(X), s(0), prod(X, fact(p(X))))
FACT(X) -> ZERO(X)
FACT(X) -> PROD(X, fact(p(X)))
FACT(X) -> FACT(p(X))
FACT(X) -> P(X)
ADD(s(X), Y) -> ADD(X, Y)
PROD(s(X), Y) -> ADD(Y, prod(X, Y))
PROD(s(X), Y) -> PROD(X, Y)

Furthermore, R contains three SCCs.

     ↳DPs

       →DP Problem 1

         ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

ADD(s(X), Y) -> ADD(X, Y)

Rules:

fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Strategy:

innermost

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

ADD(s(X), Y) -> ADD(X, Y)

one new Dependency Pair is created:

ADD(s(s(X'')), Y'') -> ADD(s(X''), Y'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳Forward Instantiation Transformation

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

ADD(s(s(X'')), Y'') -> ADD(s(X''), Y'')

Rules:

fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Strategy:

innermost

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

ADD(s(s(X'')), Y'') -> ADD(s(X''), Y'')

one new Dependency Pair is created:

ADD(s(s(s(X''''))), Y'''') -> ADD(s(s(X'''')), Y'''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 5

                 ↳Polynomial Ordering

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

ADD(s(s(s(X''''))), Y'''') -> ADD(s(s(X'''')), Y'''')

Rules:

fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

ADD(s(s(s(X''''))), Y'''') -> ADD(s(s(X'''')), Y'''')

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(ADD(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

           →DP Problem 4

             ↳FwdInst

...

               →DP Problem 6

                 ↳Dependency Graph

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining

Dependency Pair:

Rules:

fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pair:

PROD(s(X), Y) -> PROD(X, Y)

Rules:

fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Strategy:

innermost

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

PROD(s(X), Y) -> PROD(X, Y)

one new Dependency Pair is created:

PROD(s(s(X'')), Y'') -> PROD(s(X''), Y'')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳Forward Instantiation Transformation

       →DP Problem 3

         ↳Remaining

Dependency Pair:

PROD(s(s(X'')), Y'') -> PROD(s(X''), Y'')

Rules:

fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Strategy:

innermost

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

PROD(s(s(X'')), Y'') -> PROD(s(X''), Y'')

one new Dependency Pair is created:

PROD(s(s(s(X''''))), Y'''') -> PROD(s(s(X'''')), Y'''')

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 8

                 ↳Polynomial Ordering

       →DP Problem 3

         ↳Remaining

Dependency Pair:

PROD(s(s(s(X''''))), Y'''') -> PROD(s(s(X'''')), Y'''')

Rules:

fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Strategy:

innermost

The following dependency pair can be strictly oriented:

PROD(s(s(s(X''''))), Y'''') -> PROD(s(s(X'''')), Y'''')

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(s(x₁)) = 1 + x₁

POL(PROD(x₁, x₂)) = 1 + x₁

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

           →DP Problem 7

             ↳FwdInst

...

               →DP Problem 9

                 ↳Dependency Graph

       →DP Problem 3

         ↳Remaining

Dependency Pair:

Rules:

fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳FwdInst

       →DP Problem 2

         ↳FwdInst

       →DP Problem 3

         ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pair:

FACT(X) -> FACT(p(X))

Rules:

fact(X) -> if(zero(X), s(0), prod(X, fact(p(X))))
add(0, X) -> X
add(s(X), Y) -> s(add(X, Y))
prod(0, X) -> 0
prod(s(X), Y) -> add(Y, prod(X, Y))
if(true, X, Y) -> X
if(false, X, Y) -> Y
zero(0) -> true
zero(s(X)) -> false
p(s(X)) -> X

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes

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

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