(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following 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

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following 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

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

Q DP problem:
The TRS P consists of the following rules:

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)

The TRS R consists of the following 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

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following 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

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ADD(s(X), Y) → ADD(X, Y)

R is empty.
The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

(11) Obligation:

Q DP problem:
The TRS P consists of the following rules:

ADD(s(X), Y) → ADD(X, Y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • ADD(s(X), Y) → ADD(X, Y)
    The graph contains the following edges 1 > 1, 2 >= 2

(13) YES

(14) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROD(s(X), Y) → PROD(X, Y)

The TRS R consists of the following 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

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROD(s(X), Y) → PROD(X, Y)

R is empty.
The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

(18) Obligation:

Q DP problem:
The TRS P consists of the following rules:

PROD(s(X), Y) → PROD(X, Y)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

  • PROD(s(X), Y) → PROD(X, Y)
    The graph contains the following edges 1 > 1, 2 >= 2

(20) YES

(21) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FACT(X) → FACT(p(X))

The TRS R consists of the following 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

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FACT(X) → FACT(p(X))

The TRS R consists of the following rules:

p(s(X)) → X

The set Q consists of the following terms:

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))
p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

fact(x0)
add(0, x0)
add(s(x0), x1)
prod(0, x0)
prod(s(x0), x1)
if(true, x0, x1)
if(false, x0, x1)
zero(0)
zero(s(x0))

(25) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FACT(X) → FACT(p(X))

The TRS R consists of the following rules:

p(s(X)) → X

The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(26) UsableRulesReductionPairsProof (EQUIVALENT transformation)

By using the usable rules with reduction pair processor [LPAR04] with a polynomial ordering [POLO], all dependency pairs and the corresponding usable rules [FROCOS05] can be oriented non-strictly. All non-usable rules are removed, and those dependency pairs and usable rules that have been oriented strictly or contain non-usable symbols in their left-hand side are removed as well.

No dependency pairs are removed.

The following rules are removed from R:

p(s(X)) → X
Used ordering: POLO with Polynomial interpretation [POLO]:

POL(FACT(x1)) = 2·x1   
POL(p(x1)) = x1   
POL(s(x1)) = x1   

(27) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FACT(X) → FACT(p(X))

R is empty.
The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(28) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule FACT(X) → FACT(p(X)) we obtained the following new rules [LPAR04]:

FACT(p(z0)) → FACT(p(p(z0)))

(29) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FACT(p(z0)) → FACT(p(p(z0)))

R is empty.
The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(30) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule FACT(p(z0)) → FACT(p(p(z0))) we obtained the following new rules [LPAR04]:

FACT(p(p(z0))) → FACT(p(p(p(z0))))

(31) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FACT(p(p(z0))) → FACT(p(p(p(z0))))

R is empty.
The set Q consists of the following terms:

p(s(x0))

We have to consider all minimal (P,Q,R)-chains.

(32) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as they contain symbols which do neither occur in P nor in R.[THIEMANN].

p(s(x0))

(33) Obligation:

Q DP problem:
The TRS P consists of the following rules:

FACT(p(p(z0))) → FACT(p(p(p(z0))))

R is empty.
Q is empty.
We have to consider all (P,Q,R)-chains.

(34) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = FACT(p(p(z0))) evaluates to t =FACT(p(p(p(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [z0 / p(z0)]
  • Semiunifier: [ ]




Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from FACT(p(p(z0))) to FACT(p(p(p(z0)))).



(35) NO