(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
(1) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(activate(x1)) = x1
POL(and(x1, x2)) = 1 + 2·x1 + 2·x2
POL(cons(x1, x2)) = 2·x1 + 2·x2
POL(length(x1)) = 2·x1
POL(n__zeros) = 0
POL(nil) = 0
POL(s(x1)) = 2·x1
POL(tt) = 0
POL(zeros) = 0
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
and(tt, X) → activate(X)
(2) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
(3) QTRSRRRProof (EQUIVALENT transformation)
Used ordering:
Polynomial interpretation [POLO]:
POL(0) = 0
POL(activate(x1)) = 1 + x1
POL(cons(x1, x2)) = 1 + x1 + x2
POL(length(x1)) = 1 + x1
POL(n__zeros) = 0
POL(nil) = 0
POL(s(x1)) = x1
POL(zeros) = 1
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:
length(nil) → 0
zeros → n__zeros
activate(X) → X
(4) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
Q is empty.
(5) Overlay + Local Confluence (EQUIVALENT transformation)
The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.
(6) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
activate(n__zeros)
(7) DependencyPairsProof (EQUIVALENT transformation)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
LENGTH(cons(N, L)) → ACTIVATE(L)
ACTIVATE(n__zeros) → ZEROS
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
activate(n__zeros)
We have to consider all minimal (P,Q,R)-chains.
(9) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
(10) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
activate(n__zeros)
We have to consider all minimal (P,Q,R)-chains.
(11) 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.
(12) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
activate(n__zeros)
We have to consider all minimal (P,Q,R)-chains.
(13) 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].
length(cons(x0, x1))
(14) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
activate(n__zeros)
We have to consider all minimal (P,Q,R)-chains.
(15) Narrowing (EQUIVALENT transformation)
By narrowing [LPAR04] the rule
LENGTH(
cons(
N,
L)) →
LENGTH(
activate(
L)) at position [0] we obtained the following new rules [LPAR04]:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
(16) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
The TRS R consists of the following rules:
activate(n__zeros) → zeros
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
activate(n__zeros)
We have to consider all minimal (P,Q,R)-chains.
(17) 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.
(18) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
activate(n__zeros)
We have to consider all minimal (P,Q,R)-chains.
(19) 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].
activate(n__zeros)
(20) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
(21) Rewriting (EQUIVALENT transformation)
By rewriting [LPAR04] the rule
LENGTH(
cons(
y0,
n__zeros)) →
LENGTH(
zeros) at position [0] we obtained the following new rules [LPAR04]:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
(22) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
(23) 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.
(24) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
R is empty.
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
(25) 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].
zeros
(26) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(27) Instantiation (EQUIVALENT transformation)
By instantiating [LPAR04] the rule
LENGTH(
cons(
y0,
n__zeros)) →
LENGTH(
cons(
0,
n__zeros)) we obtained the following new rules [LPAR04]:
LENGTH(cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
(28) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(29) 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 =
LENGTH(
cons(
0,
n__zeros)) evaluates to t =
LENGTH(
cons(
0,
n__zeros))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Matcher: [ ]
- Semiunifier: [ ]
Rewriting sequenceThe DP semiunifies directly so there is only one rewrite step from LENGTH(cons(0, n__zeros)) to LENGTH(cons(0, n__zeros)).
(30) FALSE