Termination w.r.t. Q of the following Term Rewriting System could be disproven:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
active(and(tt, X)) → mark(X)
active(length(nil)) → mark(0)
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(active(x1)) = x1
POL(and(x1, x2)) = 1 + 2·x1 + x2
POL(cons(x1, x2)) = 2·x1 + x2
POL(length(x1)) = 2 + x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
POL(take(x1, x2)) = 2·x1 + 2·x2
POL(tt) = 0
POL(zeros) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(0, IL)) → mark(nil)
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
active(take(0, IL)) → mark(nil)
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(active(x1)) = x1
POL(and(x1, x2)) = x1 + x2
POL(cons(x1, x2)) = 2·x1 + x2
POL(length(x1)) = 2·x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
POL(take(x1, x2)) = 1 + 2·x1 + 2·x2
POL(tt) = 0
POL(zeros) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
CONS(mark(X1), X2) → CONS(X1, X2)
MARK(take(X1, X2)) → MARK(X2)
ACTIVE(length(cons(N, L))) → S(length(L))
MARK(tt) → ACTIVE(tt)
TAKE(X1, active(X2)) → TAKE(X1, X2)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
ACTIVE(length(cons(N, L))) → LENGTH(L)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(zeros) → CONS(0, zeros)
MARK(take(X1, X2)) → MARK(X1)
LENGTH(mark(X)) → LENGTH(X)
MARK(take(X1, X2)) → TAKE(mark(X1), mark(X2))
LENGTH(active(X)) → LENGTH(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(and(X1, X2)) → AND(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(take(s(M), cons(N, IL))) → CONS(N, take(M, IL))
TAKE(active(X1), X2) → TAKE(X1, X2)
ACTIVE(take(s(M), cons(N, IL))) → TAKE(M, IL)
MARK(length(X)) → ACTIVE(length(mark(X)))
AND(X1, active(X2)) → AND(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
MARK(zeros) → ACTIVE(zeros)
MARK(0) → ACTIVE(0)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(length(X)) → LENGTH(mark(X))
TAKE(mark(X1), X2) → TAKE(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(nil) → ACTIVE(nil)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
CONS(mark(X1), X2) → CONS(X1, X2)
MARK(take(X1, X2)) → MARK(X2)
ACTIVE(length(cons(N, L))) → S(length(L))
MARK(tt) → ACTIVE(tt)
TAKE(X1, active(X2)) → TAKE(X1, X2)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
CONS(X1, mark(X2)) → CONS(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
ACTIVE(length(cons(N, L))) → LENGTH(L)
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
CONS(X1, active(X2)) → CONS(X1, X2)
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(and(X1, X2)) → MARK(X1)
ACTIVE(zeros) → CONS(0, zeros)
MARK(take(X1, X2)) → MARK(X1)
LENGTH(mark(X)) → LENGTH(X)
MARK(take(X1, X2)) → TAKE(mark(X1), mark(X2))
LENGTH(active(X)) → LENGTH(X)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
S(active(X)) → S(X)
S(mark(X)) → S(X)
MARK(s(X)) → S(mark(X))
MARK(cons(X1, X2)) → CONS(mark(X1), X2)
MARK(and(X1, X2)) → AND(mark(X1), X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(take(s(M), cons(N, IL))) → CONS(N, take(M, IL))
TAKE(active(X1), X2) → TAKE(X1, X2)
ACTIVE(take(s(M), cons(N, IL))) → TAKE(M, IL)
MARK(length(X)) → ACTIVE(length(mark(X)))
AND(X1, active(X2)) → AND(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
AND(mark(X1), X2) → AND(X1, X2)
MARK(zeros) → ACTIVE(zeros)
MARK(0) → ACTIVE(0)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(length(X)) → LENGTH(mark(X))
TAKE(mark(X1), X2) → TAKE(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(nil) → ACTIVE(nil)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 13 less nodes.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TAKE(X1, active(X2)) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TAKE(X1, active(X2)) → TAKE(X1, X2)
TAKE(mark(X1), X2) → TAKE(X1, X2)
TAKE(active(X1), X2) → TAKE(X1, X2)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- TAKE(X1, active(X2)) → TAKE(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- TAKE(active(X1), X2) → TAKE(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- TAKE(mark(X1), X2) → TAKE(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- TAKE(X1, mark(X2)) → TAKE(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
S(mark(X)) → S(X)
S(active(X)) → S(X)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
S(active(X)) → S(X)
S(mark(X)) → S(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- S(mark(X)) → S(X)
The graph contains the following edges 1 > 1
- S(active(X)) → S(X)
The graph contains the following edges 1 > 1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
LENGTH(mark(X)) → LENGTH(X)
LENGTH(active(X)) → LENGTH(X)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- LENGTH(mark(X)) → LENGTH(X)
The graph contains the following edges 1 > 1
- LENGTH(active(X)) → LENGTH(X)
The graph contains the following edges 1 > 1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
AND(mark(X1), X2) → AND(X1, X2)
AND(active(X1), X2) → AND(X1, X2)
AND(X1, mark(X2)) → AND(X1, X2)
AND(X1, active(X2)) → AND(X1, X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- AND(mark(X1), X2) → AND(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- AND(active(X1), X2) → AND(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- AND(X1, mark(X2)) → AND(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- AND(X1, active(X2)) → AND(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONS(X1, active(X2)) → CONS(X1, X2)
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(X1, active(X2)) → CONS(X1, X2)
CONS(X1, mark(X2)) → CONS(X1, X2)
CONS(active(X1), X2) → CONS(X1, X2)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] 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:
- CONS(X1, active(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
- CONS(mark(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- CONS(active(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- CONS(X1, mark(X2)) → CONS(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
MARK(take(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(length(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(and(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → MARK(X1)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
MARK(and(X1, X2)) → MARK(X1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(0) = 0
POL(ACTIVE(x1)) = 2·x1
POL(MARK(x1)) = 2·x1
POL(active(x1)) = x1
POL(and(x1, x2)) = 2 + 2·x1 + 2·x2
POL(cons(x1, x2)) = 2·x1 + x2
POL(length(x1)) = 2·x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
POL(take(x1, x2)) = x1 + 2·x2
POL(tt) = 0
POL(zeros) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
MARK(take(X1, X2)) → MARK(X2)
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(length(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(take(X1, X2)) → MARK(X1)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
MARK(take(X1, X2)) → MARK(X2)
MARK(take(X1, X2)) → MARK(X1)
Used ordering: POLO with Polynomial interpretation [25]:
POL(0) = 0
POL(ACTIVE(x1)) = 2·x1
POL(MARK(x1)) = 2·x1
POL(active(x1)) = x1
POL(and(x1, x2)) = 2·x1 + 2·x2
POL(cons(x1, x2)) = 2·x1 + x2
POL(length(x1)) = x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
POL(take(x1, x2)) = 2 + 2·x1 + 2·x2
POL(tt) = 0
POL(zeros) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
Q DP problem:
The TRS P consists of the following rules:
MARK(length(X)) → MARK(X)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the rule removal processor [15] with the following polynomial ordering [25], at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:
MARK(length(X)) → MARK(X)
Used ordering: POLO with Polynomial interpretation [25]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = x1
POL(active(x1)) = x1
POL(and(x1, x2)) = 2·x1 + x2
POL(cons(x1, x2)) = 2·x1 + x2
POL(length(x1)) = 1 + 2·x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
POL(take(x1, x2)) = 2·x1 + 2·x2
POL(tt) = 0
POL(zeros) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → ACTIVE(s(mark(X)))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
MARK(cons(X1, X2)) → ACTIVE(cons(mark(X1), X2))
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(and(X1, X2)) → ACTIVE(and(mark(X1), X2))
The remaining pairs can at least be oriented weakly.
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = 1
POL(active(x1)) = 0
POL(and(x1, x2)) = 0
POL(cons(x1, x2)) = 0
POL(length(x1)) = 1
POL(mark(x1)) = 0
POL(nil) = 0
POL(s(x1)) = 0
POL(take(x1, x2)) = 1
POL(tt) = 0
POL(zeros) = 1
The following usable rules [17] were oriented:
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
ACTIVE(zeros) → MARK(cons(0, zeros))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
ACTIVE(zeros) → MARK(cons(0, zeros))
The remaining pairs can at least be oriented weakly.
MARK(zeros) → ACTIVE(zeros)
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = x1
POL(active(x1)) = x1
POL(and(x1, x2)) = 0
POL(cons(x1, x2)) = x1
POL(length(x1)) = 0
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
POL(take(x1, x2)) = x2
POL(tt) = 0
POL(zeros) = 1
The following usable rules [17] were oriented:
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
mark(nil) → active(nil)
mark(0) → active(0)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(tt) → active(tt)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(zeros) → ACTIVE(zeros)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
ACTIVE(take(s(M), cons(N, IL))) → MARK(cons(N, take(M, IL)))
The remaining pairs can at least be oriented weakly.
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 0
POL(ACTIVE(x1)) = x1
POL(MARK(x1)) = x1
POL(active(x1)) = x1
POL(and(x1, x2)) = 0
POL(cons(x1, x2)) = x1
POL(length(x1)) = 0
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
POL(take(x1, x2)) = 1 + x2
POL(tt) = 0
POL(zeros) = 0
The following usable rules [17] were oriented:
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
mark(nil) → active(nil)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
mark(0) → active(0)
s(active(X)) → s(X)
s(mark(X)) → s(X)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(tt) → active(tt)
length(active(X)) → length(X)
length(mark(X)) → length(X)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(cons(X1, X2)) → MARK(X1)
MARK(length(X)) → ACTIVE(length(mark(X)))
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
MARK(cons(X1, X2)) → MARK(X1)
MARK(take(X1, X2)) → ACTIVE(take(mark(X1), mark(X2)))
The remaining pairs can at least be oriented weakly.
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(X)) → ACTIVE(length(mark(X)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 0
POL(ACTIVE(x1)) = 1
POL(MARK(x1)) = 1 + x1
POL(active(x1)) = x1
POL(and(x1, x2)) = 0
POL(cons(x1, x2)) = 1 + x1
POL(length(x1)) = 0
POL(mark(x1)) = x1
POL(nil) = 0
POL(s(x1)) = x1
POL(take(x1, x2)) = 1
POL(tt) = 0
POL(zeros) = 0
The following usable rules [17] were oriented:
s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(X)) → ACTIVE(length(mark(X)))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MARK(length(X)) → ACTIVE(length(mark(X))) at position [0] we obtained the following new rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(tt)) → ACTIVE(length(active(tt)))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(0)) → ACTIVE(length(active(0)))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(x0)) → ACTIVE(length(x0))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(tt)) → ACTIVE(length(active(tt)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(0)) → ACTIVE(length(active(0)))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MARK(length(tt)) → ACTIVE(length(active(tt))) at position [0] we obtained the following new rules:
MARK(length(tt)) → ACTIVE(length(tt))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(0)) → ACTIVE(length(active(0)))
MARK(length(tt)) → ACTIVE(length(tt))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(0)) → ACTIVE(length(active(0)))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MARK(length(0)) → ACTIVE(length(active(0))) at position [0] we obtained the following new rules:
MARK(length(0)) → ACTIVE(length(0))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MARK(length(0)) → ACTIVE(length(0))
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(nil)) → ACTIVE(length(active(nil)))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule MARK(length(nil)) → ACTIVE(length(active(nil))) at position [0] we obtained the following new rules:
MARK(length(nil)) → ACTIVE(length(nil))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(nil)) → ACTIVE(length(nil))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
MARK(length(and(x0, x1))) → ACTIVE(length(active(and(mark(x0), x1))))
The remaining pairs can at least be oriented weakly.
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( cons(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( and(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( take(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
mark(nil) → active(nil)
mark(0) → active(0)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(tt) → active(tt)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
MARK(length(length(x0))) → ACTIVE(length(active(length(mark(x0)))))
The remaining pairs can at least be oriented weakly.
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( cons(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( and(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( take(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
mark(nil) → active(nil)
mark(0) → active(0)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(tt) → active(tt)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].
The following pairs can be oriented strictly and are deleted.
MARK(length(s(x0))) → ACTIVE(length(active(s(mark(x0)))))
The remaining pairs can at least be oriented weakly.
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(s(X)) → MARK(X)
MARK(length(x0)) → ACTIVE(length(x0))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( cons(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( and(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( take(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
Matrix type:
We used a basic matrix type which is not further parametrizeable.
As matrix orders are CE-compatible, we used usable rules w.r.t. argument filtering in the order.
The following usable rules [17] were oriented:
and(X1, mark(X2)) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(mark(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
s(active(X)) → s(X)
s(mark(X)) → s(X)
length(active(X)) → length(X)
length(mark(X)) → length(X)
mark(nil) → active(nil)
mark(0) → active(0)
active(zeros) → mark(cons(0, zeros))
mark(length(X)) → active(length(mark(X)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
mark(and(X1, X2)) → active(and(mark(X1), X2))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(s(X)) → active(s(mark(X)))
active(length(cons(N, L))) → mark(s(length(L)))
mark(tt) → active(tt)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ RuleRemovalProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
The TRS P consists of the following rules:
MARK(length(take(x0, x1))) → ACTIVE(length(active(take(mark(x0), mark(x1)))))
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
MARK(length(cons(x0, x1))) → ACTIVE(length(active(cons(mark(x0), x1))))
MARK(s(X)) → MARK(X)
ACTIVE(length(cons(N, L))) → MARK(s(length(L)))
MARK(length(x0)) → ACTIVE(length(x0))
The TRS R consists of the following rules:
active(zeros) → mark(cons(0, zeros))
active(length(cons(N, L))) → mark(s(length(L)))
active(take(s(M), cons(N, IL))) → mark(cons(N, take(M, IL)))
mark(zeros) → active(zeros)
mark(cons(X1, X2)) → active(cons(mark(X1), X2))
mark(0) → active(0)
mark(and(X1, X2)) → active(and(mark(X1), X2))
mark(tt) → active(tt)
mark(length(X)) → active(length(mark(X)))
mark(nil) → active(nil)
mark(s(X)) → active(s(mark(X)))
mark(take(X1, X2)) → active(take(mark(X1), mark(X2)))
cons(mark(X1), X2) → cons(X1, X2)
cons(X1, mark(X2)) → cons(X1, X2)
cons(active(X1), X2) → cons(X1, X2)
cons(X1, active(X2)) → cons(X1, X2)
and(mark(X1), X2) → and(X1, X2)
and(X1, mark(X2)) → and(X1, X2)
and(active(X1), X2) → and(X1, X2)
and(X1, active(X2)) → and(X1, X2)
length(mark(X)) → length(X)
length(active(X)) → length(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
take(mark(X1), X2) → take(X1, X2)
take(X1, mark(X2)) → take(X1, X2)
take(active(X1), X2) → take(X1, X2)
take(X1, active(X2)) → take(X1, X2)
s = ACTIVE(length(active(zeros))) evaluates to t =ACTIVE(length(active(zeros)))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
ACTIVE(length(active(zeros))) → ACTIVE(length(mark(cons(0, zeros))))
with rule active(zeros) → mark(cons(0, zeros)) at position [0,0] and matcher [ ]
ACTIVE(length(mark(cons(0, zeros)))) → ACTIVE(length(cons(0, zeros)))
with rule length(mark(X)) → length(X) at position [0] and matcher [X / cons(0, zeros)]
ACTIVE(length(cons(0, zeros))) → MARK(s(length(zeros)))
with rule ACTIVE(length(cons(N, L))) → MARK(s(length(L))) at position [] and matcher [N / 0, L / zeros]
MARK(s(length(zeros))) → MARK(length(zeros))
with rule MARK(s(X)) → MARK(X) at position [] and matcher [X / length(zeros)]
MARK(length(zeros)) → ACTIVE(length(active(zeros)))
with rule MARK(length(zeros)) → ACTIVE(length(active(zeros)))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.