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

         ↳Usable Rules (Innermost)

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

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

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

           →DP Problem 4

             ↳Size-Change Principle

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

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

As we are in the innermost case, we can delete all 10 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

           →DP Problem 5

             ↳Size-Change Principle

       →DP Problem 3

         ↳UsableRules

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:

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

As we are in the innermost case, we can delete all 9 non-usable-rules.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳Modular Removal of Rules

Dependency Pair:

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

Rule:

p(s(X)) -> X

Strategy:

innermost

We have the following set of usable rules:

p(s(X)) -> X

To remove rules and DPs from this DP problem we used the following monotonic and C_E-compatible order: Polynomial ordering.
Polynomial interpretation:

POL(FACT(x₁)) = x₁

POL(s(x₁)) = x₁

POL(p(x₁)) = x₁

We have the following set D of usable symbols: {FACT, p}
No Dependency Pairs can be deleted.
The following rules can be deleted as they contain symbols in their lhs which do not occur in D:

p(s(X)) -> X

The result of this processor delivers one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳UsableRules

       →DP Problem 2

         ↳UsableRules

       →DP Problem 3

         ↳UsableRules

           →DP Problem 6

             ↳MRR

...

               →DP Problem 7

                 ↳Non Termination

Dependency Pair:

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

Rule:

none

Strategy:

innermost

Found an infinite P-chain over R:
P =

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

R = none

s = FACT(X)
evaluates to t =FACT(p(X))

Thus, s starts an infinite chain as s matches t.

Innermost Non-Termination of R could be shown.
Duration:
0:01 minutes

POL(FACT(x₁))	= x₁
POL(s(x₁))	= x₁
POL(p(x₁))	= x₁