Termination w.r.t. Q of the following Term Rewriting System could not be shown:
Q restricted rewrite system:
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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:
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
FROM(mark(X)) → FROM(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → TAKE(X1, active(X2))
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
2ND(mark(X)) → 2ND(X)
PROPER(head(X)) → PROPER(X)
HEAD(mark(X)) → HEAD(X)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
2ND(ok(X)) → 2ND(X)
PROPER(2nd(X)) → PROPER(X)
ACTIVE(2nd(cons(X, XS))) → HEAD(XS)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
PROPER(from(X)) → FROM(proper(X))
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X2)
ACTIVE(take(X1, X2)) → TAKE(active(X1), X2)
PROPER(s(X)) → S(proper(X))
ACTIVE(take(s(N), cons(X, XS))) → TAKE(N, XS)
ACTIVE(2nd(X)) → 2ND(active(X))
ACTIVE(head(X)) → HEAD(active(X))
ACTIVE(s(X)) → ACTIVE(X)
TAKE(mark(X1), X2) → TAKE(X1, X2)
FROM(ok(X)) → FROM(X)
ACTIVE(from(X)) → FROM(s(X))
S(ok(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
PROPER(take(X1, X2)) → TAKE(proper(X1), proper(X2))
TOP(mark(X)) → PROPER(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
TOP(ok(X)) → ACTIVE(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
SEL(X1, mark(X2)) → SEL(X1, X2)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
ACTIVE(from(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
S(mark(X)) → S(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(take(s(N), cons(X, XS))) → CONS(X, take(N, XS))
PROPER(head(X)) → HEAD(proper(X))
PROPER(2nd(X)) → 2ND(proper(X))
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → S(X)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X1)
ACTIVE(head(X)) → ACTIVE(X)
TOP(mark(X)) → TOP(proper(X))
HEAD(ok(X)) → HEAD(X)
ACTIVE(sel(s(N), cons(X, XS))) → SEL(N, XS)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(s(X)) → S(active(X))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(sel(X1, X2)) → SEL(active(X1), X2)
FROM(mark(X)) → FROM(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → TAKE(X1, active(X2))
ACTIVE(cons(X1, X2)) → CONS(active(X1), X2)
2ND(mark(X)) → 2ND(X)
PROPER(head(X)) → PROPER(X)
HEAD(mark(X)) → HEAD(X)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
2ND(ok(X)) → 2ND(X)
PROPER(2nd(X)) → PROPER(X)
ACTIVE(2nd(cons(X, XS))) → HEAD(XS)
PROPER(cons(X1, X2)) → PROPER(X2)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
PROPER(from(X)) → FROM(proper(X))
PROPER(sel(X1, X2)) → SEL(proper(X1), proper(X2))
PROPER(sel(X1, X2)) → PROPER(X2)
ACTIVE(take(X1, X2)) → TAKE(active(X1), X2)
PROPER(s(X)) → S(proper(X))
ACTIVE(take(s(N), cons(X, XS))) → TAKE(N, XS)
ACTIVE(2nd(X)) → 2ND(active(X))
ACTIVE(head(X)) → HEAD(active(X))
ACTIVE(s(X)) → ACTIVE(X)
TAKE(mark(X1), X2) → TAKE(X1, X2)
FROM(ok(X)) → FROM(X)
ACTIVE(from(X)) → FROM(s(X))
S(ok(X)) → S(X)
CONS(mark(X1), X2) → CONS(X1, X2)
SEL(mark(X1), X2) → SEL(X1, X2)
ACTIVE(take(X1, X2)) → ACTIVE(X2)
PROPER(take(X1, X2)) → TAKE(proper(X1), proper(X2))
TOP(mark(X)) → PROPER(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
TOP(ok(X)) → ACTIVE(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
SEL(X1, mark(X2)) → SEL(X1, X2)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
ACTIVE(from(X)) → ACTIVE(X)
TOP(ok(X)) → TOP(active(X))
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
ACTIVE(sel(X1, X2)) → SEL(X1, active(X2))
S(mark(X)) → S(X)
PROPER(cons(X1, X2)) → CONS(proper(X1), proper(X2))
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
ACTIVE(take(s(N), cons(X, XS))) → CONS(X, take(N, XS))
PROPER(head(X)) → HEAD(proper(X))
PROPER(2nd(X)) → 2ND(proper(X))
ACTIVE(2nd(X)) → ACTIVE(X)
ACTIVE(from(X)) → S(X)
TAKE(X1, mark(X2)) → TAKE(X1, X2)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X1)
ACTIVE(head(X)) → ACTIVE(X)
TOP(mark(X)) → TOP(proper(X))
HEAD(ok(X)) → HEAD(X)
ACTIVE(sel(s(N), cons(X, XS))) → SEL(N, XS)
ACTIVE(from(X)) → CONS(X, from(s(X)))
ACTIVE(from(X)) → FROM(active(X))
ACTIVE(s(X)) → S(active(X))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 10 SCCs with 25 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(X1, X2)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
SEL(mark(X1), X2) → SEL(X1, X2)
SEL(ok(X1), ok(X2)) → SEL(X1, X2)
SEL(X1, mark(X2)) → SEL(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:
- SEL(mark(X1), X2) → SEL(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- SEL(ok(X1), ok(X2)) → SEL(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2
- SEL(X1, mark(X2)) → SEL(X1, X2)
The graph contains the following edges 1 >= 1, 2 > 2
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TAKE(ok(X1), ok(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(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TAKE(ok(X1), ok(X2)) → TAKE(X1, X2)
TAKE(mark(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(ok(X1), ok(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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
2ND(ok(X)) → 2ND(X)
2ND(mark(X)) → 2ND(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
2ND(ok(X)) → 2ND(X)
2ND(mark(X)) → 2ND(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:
- 2ND(ok(X)) → 2ND(X)
The graph contains the following edges 1 > 1
- 2ND(mark(X)) → 2ND(X)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HEAD(mark(X)) → HEAD(X)
HEAD(ok(X)) → HEAD(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HEAD(mark(X)) → HEAD(X)
HEAD(ok(X)) → HEAD(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:
- HEAD(mark(X)) → HEAD(X)
The graph contains the following edges 1 > 1
- HEAD(ok(X)) → HEAD(X)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
S(ok(X)) → S(X)
S(mark(X)) → S(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
S(ok(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(ok(X)) → S(X)
The graph contains the following edges 1 > 1
- S(mark(X)) → S(X)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(ok(X1), ok(X2)) → CONS(X1, X2)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONS(mark(X1), X2) → CONS(X1, X2)
CONS(ok(X1), ok(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(mark(X1), X2) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 >= 2
- CONS(ok(X1), ok(X2)) → CONS(X1, X2)
The graph contains the following edges 1 > 1, 2 > 2
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
FROM(mark(X)) → FROM(X)
FROM(ok(X)) → FROM(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
FROM(mark(X)) → FROM(X)
FROM(ok(X)) → FROM(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:
- FROM(mark(X)) → FROM(X)
The graph contains the following edges 1 > 1
- FROM(ok(X)) → FROM(X)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(2nd(X)) → PROPER(X)
PROPER(s(X)) → PROPER(X)
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(head(X)) → PROPER(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER(cons(X1, X2)) → PROPER(X1)
PROPER(s(X)) → PROPER(X)
PROPER(2nd(X)) → PROPER(X)
PROPER(take(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X1)
PROPER(sel(X1, X2)) → PROPER(X2)
PROPER(cons(X1, X2)) → PROPER(X2)
PROPER(take(X1, X2)) → PROPER(X2)
PROPER(from(X)) → PROPER(X)
PROPER(head(X)) → PROPER(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:
- PROPER(take(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
- PROPER(2nd(X)) → PROPER(X)
The graph contains the following edges 1 > 1
- PROPER(s(X)) → PROPER(X)
The graph contains the following edges 1 > 1
- PROPER(cons(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
- PROPER(sel(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
- PROPER(sel(X1, X2)) → PROPER(X1)
The graph contains the following edges 1 > 1
- PROPER(take(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
- PROPER(cons(X1, X2)) → PROPER(X2)
The graph contains the following edges 1 > 1
- PROPER(from(X)) → PROPER(X)
The graph contains the following edges 1 > 1
- PROPER(head(X)) → PROPER(X)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(2nd(X)) → ACTIVE(X)
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
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
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QDPSizeChangeProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE(take(X1, X2)) → ACTIVE(X2)
ACTIVE(s(X)) → ACTIVE(X)
ACTIVE(sel(X1, X2)) → ACTIVE(X1)
ACTIVE(head(X)) → ACTIVE(X)
ACTIVE(cons(X1, X2)) → ACTIVE(X1)
ACTIVE(take(X1, X2)) → ACTIVE(X1)
ACTIVE(sel(X1, X2)) → ACTIVE(X2)
ACTIVE(from(X)) → ACTIVE(X)
ACTIVE(2nd(X)) → ACTIVE(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:
- ACTIVE(take(X1, X2)) → ACTIVE(X2)
The graph contains the following edges 1 > 1
- ACTIVE(sel(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
- ACTIVE(s(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
- ACTIVE(head(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
- ACTIVE(cons(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
- ACTIVE(sel(X1, X2)) → ACTIVE(X2)
The graph contains the following edges 1 > 1
- ACTIVE(take(X1, X2)) → ACTIVE(X1)
The graph contains the following edges 1 > 1
- ACTIVE(from(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
- ACTIVE(2nd(X)) → ACTIVE(X)
The graph contains the following edges 1 > 1
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
from(mark(X)) → mark(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
s(mark(X)) → mark(s(X))
head(mark(X)) → mark(head(X))
2nd(mark(X)) → mark(2nd(X))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
from(ok(X)) → ok(from(X))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(ok(X)) → ok(s(X))
head(ok(X)) → ok(head(X))
2nd(ok(X)) → ok(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
top(mark(X)) → top(proper(X))
top(ok(X)) → top(active(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the usable rules with reduction pair processor [15] with a polynomial ordering [25], all dependency pairs and the corresponding usable rules [17] 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.
No rules are removed from R.
Used ordering: POLO with Polynomial interpretation [25]:
POL(0) = 0
POL(2nd(x1)) = x1
POL(TOP(x1)) = 2·x1
POL(active(x1)) = 2·x1
POL(cons(x1, x2)) = x1 + x2
POL(from(x1)) = x1
POL(head(x1)) = x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(ok(x1)) = 2·x1
POL(proper(x1)) = x1
POL(s(x1)) = x1
POL(sel(x1, x2)) = 2·x1 + x2
POL(take(x1, x2)) = 2·x1 + 2·x2
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(X)) → TOP(proper(X))
TOP(ok(X)) → TOP(active(X))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(mark(X)) → TOP(proper(X)) at position [0] we obtained the following new rules:
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(mark(nil)) → TOP(ok(nil))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(mark(0)) → TOP(ok(0))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(nil)) → TOP(ok(nil))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(X)) → TOP(active(X))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule TOP(ok(X)) → TOP(active(X)) at position [0] we obtained the following new rules:
TOP(ok(2nd(cons(x0, x1)))) → TOP(mark(head(x1)))
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(ok(take(0, x0))) → TOP(mark(nil))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(2nd(cons(x0, x1)))) → TOP(mark(head(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(mark(nil)) → TOP(ok(nil))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(take(0, x0))) → TOP(mark(nil))
TOP(mark(0)) → TOP(ok(0))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(ok(head(cons(x0, x1)))) → TOP(mark(x0))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(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 3 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(2nd(cons(x0, x1)))) → TOP(mark(head(x1)))
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(head(cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(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.
TOP(ok(sel(0, cons(x0, x1)))) → TOP(mark(x0))
The remaining pairs can at least be oriented weakly.
TOP(ok(2nd(cons(x0, x1)))) → TOP(mark(head(x1)))
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(head(cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 0
POL(2nd(x1)) = x1
POL(TOP(x1)) = x1
POL(active(x1)) = x1
POL(cons(x1, x2)) = x1 + x2
POL(from(x1)) = x1
POL(head(x1)) = x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(ok(x1)) = x1
POL(proper(x1)) = x1
POL(s(x1)) = 0
POL(sel(x1, x2)) = 1 + x2
POL(take(x1, x2)) = x2
The following usable rules [17] were oriented:
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
head(ok(X)) → ok(head(X))
head(mark(X)) → mark(head(X))
2nd(ok(X)) → ok(2nd(X))
2nd(mark(X)) → mark(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
active(sel(X1, X2)) → sel(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(take(X1, X2)) → take(active(X1), X2)
active(2nd(X)) → 2nd(active(X))
active(head(X)) → head(active(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(2nd(cons(x0, x1)))) → TOP(mark(head(x1)))
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(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.
TOP(ok(2nd(cons(x0, x1)))) → TOP(mark(head(x1)))
The remaining pairs can at least be oriented weakly.
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 0
POL(2nd(x1)) = 1 + x1
POL(TOP(x1)) = x1
POL(active(x1)) = x1
POL(cons(x1, x2)) = x1 + x2
POL(from(x1)) = x1
POL(head(x1)) = x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(ok(x1)) = x1
POL(proper(x1)) = x1
POL(s(x1)) = 0
POL(sel(x1, x2)) = x2
POL(take(x1, x2)) = x2
The following usable rules [17] were oriented:
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
head(ok(X)) → ok(head(X))
head(mark(X)) → mark(head(X))
2nd(ok(X)) → ok(2nd(X))
2nd(mark(X)) → mark(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
active(sel(X1, X2)) → sel(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(take(X1, X2)) → take(active(X1), X2)
active(2nd(X)) → 2nd(active(X))
active(head(X)) → head(active(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(cons(x0, x1)))) → TOP(mark(x0))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(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.
TOP(ok(head(cons(x0, x1)))) → TOP(mark(x0))
The remaining pairs can at least be oriented weakly.
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 0
POL(2nd(x1)) = 1 + x1
POL(TOP(x1)) = x1
POL(active(x1)) = x1
POL(cons(x1, x2)) = x1 + x2
POL(from(x1)) = x1
POL(head(x1)) = 1 + x1
POL(mark(x1)) = x1
POL(nil) = 0
POL(ok(x1)) = x1
POL(proper(x1)) = x1
POL(s(x1)) = 0
POL(sel(x1, x2)) = x2
POL(take(x1, x2)) = x2
The following usable rules [17] were oriented:
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
head(ok(X)) → ok(head(X))
head(mark(X)) → mark(head(X))
2nd(ok(X)) → ok(2nd(X))
2nd(mark(X)) → mark(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
active(sel(X1, X2)) → sel(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(take(X1, X2)) → take(active(X1), X2)
active(2nd(X)) → 2nd(active(X))
active(head(X)) → head(active(X))
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(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.
TOP(ok(take(s(x0), cons(x1, x2)))) → TOP(mark(cons(x1, take(x0, x2))))
The remaining pairs can at least be oriented weakly.
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
Used ordering: Polynomial interpretation [25]:
POL(0) = 0
POL(2nd(x1)) = 0
POL(TOP(x1)) = x1
POL(active(x1)) = 0
POL(cons(x1, x2)) = 0
POL(from(x1)) = 0
POL(head(x1)) = 0
POL(mark(x1)) = x1
POL(nil) = 0
POL(ok(x1)) = x1
POL(proper(x1)) = 0
POL(s(x1)) = 0
POL(sel(x1, x2)) = 1
POL(take(x1, x2)) = 1
The following usable rules [17] were oriented:
head(ok(X)) → ok(head(X))
head(mark(X)) → mark(head(X))
2nd(ok(X)) → ok(2nd(X))
2nd(mark(X)) → mark(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(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.
TOP(ok(from(x0))) → TOP(mark(cons(x0, from(s(x0)))))
The remaining pairs can at least be oriented weakly.
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( sel(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( take(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
| M( cons(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:
head(ok(X)) → ok(head(X))
head(mark(X)) → mark(head(X))
2nd(ok(X)) → ok(2nd(X))
2nd(mark(X)) → mark(2nd(X))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ UsableRulesReductionPairsProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ QDPOrderProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TOP(ok(from(x0))) → TOP(from(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(x0, active(x1)))
TOP(mark(from(x0))) → TOP(from(proper(x0)))
TOP(mark(s(x0))) → TOP(s(proper(x0)))
TOP(ok(2nd(x0))) → TOP(2nd(active(x0)))
TOP(ok(s(x0))) → TOP(s(active(x0)))
TOP(ok(sel(x0, x1))) → TOP(sel(active(x0), x1))
TOP(mark(sel(x0, x1))) → TOP(sel(proper(x0), proper(x1)))
TOP(mark(head(x0))) → TOP(head(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(x0, active(x1)))
TOP(ok(head(x0))) → TOP(head(active(x0)))
TOP(ok(sel(s(x0), cons(x1, x2)))) → TOP(mark(sel(x0, x2)))
TOP(ok(cons(x0, x1))) → TOP(cons(active(x0), x1))
TOP(mark(cons(x0, x1))) → TOP(cons(proper(x0), proper(x1)))
TOP(mark(2nd(x0))) → TOP(2nd(proper(x0)))
TOP(ok(take(x0, x1))) → TOP(take(active(x0), x1))
TOP(mark(take(x0, x1))) → TOP(take(proper(x0), proper(x1)))
The TRS R consists of the following rules:
active(from(X)) → mark(cons(X, from(s(X))))
active(head(cons(X, XS))) → mark(X)
active(2nd(cons(X, XS))) → mark(head(XS))
active(take(0, XS)) → mark(nil)
active(take(s(N), cons(X, XS))) → mark(cons(X, take(N, XS)))
active(sel(0, cons(X, XS))) → mark(X)
active(sel(s(N), cons(X, XS))) → mark(sel(N, XS))
active(from(X)) → from(active(X))
active(cons(X1, X2)) → cons(active(X1), X2)
active(s(X)) → s(active(X))
active(head(X)) → head(active(X))
active(2nd(X)) → 2nd(active(X))
active(take(X1, X2)) → take(active(X1), X2)
active(take(X1, X2)) → take(X1, active(X2))
active(sel(X1, X2)) → sel(active(X1), X2)
active(sel(X1, X2)) → sel(X1, active(X2))
sel(mark(X1), X2) → mark(sel(X1, X2))
sel(X1, mark(X2)) → mark(sel(X1, X2))
sel(ok(X1), ok(X2)) → ok(sel(X1, X2))
take(mark(X1), X2) → mark(take(X1, X2))
take(X1, mark(X2)) → mark(take(X1, X2))
take(ok(X1), ok(X2)) → ok(take(X1, X2))
2nd(mark(X)) → mark(2nd(X))
2nd(ok(X)) → ok(2nd(X))
head(mark(X)) → mark(head(X))
head(ok(X)) → ok(head(X))
s(mark(X)) → mark(s(X))
s(ok(X)) → ok(s(X))
cons(mark(X1), X2) → mark(cons(X1, X2))
cons(ok(X1), ok(X2)) → ok(cons(X1, X2))
from(mark(X)) → mark(from(X))
from(ok(X)) → ok(from(X))
proper(from(X)) → from(proper(X))
proper(cons(X1, X2)) → cons(proper(X1), proper(X2))
proper(s(X)) → s(proper(X))
proper(head(X)) → head(proper(X))
proper(2nd(X)) → 2nd(proper(X))
proper(take(X1, X2)) → take(proper(X1), proper(X2))
proper(0) → ok(0)
proper(nil) → ok(nil)
proper(sel(X1, X2)) → sel(proper(X1), proper(X2))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.