(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
fact(X) → if(zero(X), n__s(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
activate(X) → X
Q is empty.
(1) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(2) Obligation:
Q DP problem:
The TRS P consists of the following rules:
FACT(X) → IF(zero(X), n__s(0), n__prod(X, fact(p(X))))
FACT(X) → ZERO(X)
FACT(X) → FACT(p(X))
FACT(X) → P(X)
ADD(s(X), Y) → S(add(X, Y))
ADD(s(X), Y) → ADD(X, Y)
PROD(s(X), Y) → ADD(Y, prod(X, Y))
PROD(s(X), Y) → PROD(X, Y)
IF(true, X, Y) → ACTIVATE(X)
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__s(X)) → S(X)
ACTIVATE(n__prod(X1, X2)) → PROD(X1, X2)
The TRS R consists of the following rules:
fact(X) → if(zero(X), n__s(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(3) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 9 less nodes.
(4) Complex Obligation (AND)
(5) 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), n__s(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(6) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(7) 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.
(8) 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
(9) YES
(10) 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), n__s(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(11) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(12) 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.
(13) 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
(14) YES
(15) 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), n__s(0), n__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) → activate(X)
if(false, X, Y) → activate(Y)
zero(0) → true
zero(s(X)) → false
p(s(X)) → X
s(X) → n__s(X)
prod(X1, X2) → n__prod(X1, X2)
activate(n__s(X)) → s(X)
activate(n__prod(X1, X2)) → prod(X1, X2)
activate(X) → X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(16) UsableRulesProof (EQUIVALENT transformation)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
(17) 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
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(18) MNOCProof (EQUIVALENT transformation)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
(19) 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.
(20) 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
(21) 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.
(22) 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)))
(23) 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.
(24) 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))))
(25) 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.
(26) 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))
(27) 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.
(28) 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 sequenceThe DP semiunifies directly so there is only one rewrite step from FACT(p(p(z0))) to FACT(p(p(p(z0)))).
(29) NO