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:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
PROPER1(head1(X)) -> PROPER1(X)
S1(mark1(X)) -> S1(X)
PROPER1(adx1(X)) -> PROPER1(X)
INCR1(mark1(X)) -> INCR1(X)
ACTIVE1(adx1(cons2(X, L))) -> ADX1(L)
PROPER1(incr1(X)) -> INCR1(proper1(X))
PROPER1(incr1(X)) -> PROPER1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
ACTIVE1(tail1(X)) -> TAIL1(active1(X))
HEAD1(mark1(X)) -> HEAD1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
INCR1(ok1(X)) -> INCR1(X)
ADX1(mark1(X)) -> ADX1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(tail1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(tail1(X)) -> TAIL1(proper1(X))
ACTIVE1(adx1(X)) -> ADX1(active1(X))
ACTIVE1(tail1(X)) -> ACTIVE1(X)
HEAD1(ok1(X)) -> HEAD1(X)
ACTIVE1(zeros) -> CONS2(0, zeros)
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(adx1(cons2(X, L))) -> CONS2(X, adx1(L))
TAIL1(ok1(X)) -> TAIL1(X)
ACTIVE1(incr1(cons2(X, L))) -> INCR1(L)
PROPER1(adx1(X)) -> ADX1(proper1(X))
ACTIVE1(head1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(cons2(X, L))) -> S1(X)
PROPER1(s1(X)) -> S1(proper1(X))
ACTIVE1(head1(X)) -> HEAD1(active1(X))
ACTIVE1(s1(X)) -> ACTIVE1(X)
S1(ok1(X)) -> S1(X)
ACTIVE1(nats) -> ADX1(zeros)
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
TAIL1(mark1(X)) -> TAIL1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
PROPER1(head1(X)) -> HEAD1(proper1(X))
ACTIVE1(adx1(cons2(X, L))) -> INCR1(cons2(X, adx1(L)))
ACTIVE1(incr1(X)) -> INCR1(active1(X))
ACTIVE1(incr1(cons2(X, L))) -> CONS2(s1(X), incr1(L))
ADX1(ok1(X)) -> ADX1(X)
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(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:
PROPER1(head1(X)) -> PROPER1(X)
S1(mark1(X)) -> S1(X)
PROPER1(adx1(X)) -> PROPER1(X)
INCR1(mark1(X)) -> INCR1(X)
ACTIVE1(adx1(cons2(X, L))) -> ADX1(L)
PROPER1(incr1(X)) -> INCR1(proper1(X))
PROPER1(incr1(X)) -> PROPER1(X)
TOP1(mark1(X)) -> TOP1(proper1(X))
ACTIVE1(tail1(X)) -> TAIL1(active1(X))
HEAD1(mark1(X)) -> HEAD1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
INCR1(ok1(X)) -> INCR1(X)
ADX1(mark1(X)) -> ADX1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(tail1(X)) -> PROPER1(X)
TOP1(ok1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
ACTIVE1(s1(X)) -> S1(active1(X))
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(tail1(X)) -> TAIL1(proper1(X))
ACTIVE1(adx1(X)) -> ADX1(active1(X))
ACTIVE1(tail1(X)) -> ACTIVE1(X)
HEAD1(ok1(X)) -> HEAD1(X)
ACTIVE1(zeros) -> CONS2(0, zeros)
TOP1(mark1(X)) -> PROPER1(X)
ACTIVE1(adx1(cons2(X, L))) -> CONS2(X, adx1(L))
TAIL1(ok1(X)) -> TAIL1(X)
ACTIVE1(incr1(cons2(X, L))) -> INCR1(L)
PROPER1(adx1(X)) -> ADX1(proper1(X))
ACTIVE1(head1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(cons2(X, L))) -> S1(X)
PROPER1(s1(X)) -> S1(proper1(X))
ACTIVE1(head1(X)) -> HEAD1(active1(X))
ACTIVE1(s1(X)) -> ACTIVE1(X)
S1(ok1(X)) -> S1(X)
ACTIVE1(nats) -> ADX1(zeros)
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
ACTIVE1(cons2(X1, X2)) -> CONS2(active1(X1), X2)
TAIL1(mark1(X)) -> TAIL1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
PROPER1(head1(X)) -> HEAD1(proper1(X))
ACTIVE1(adx1(cons2(X, L))) -> INCR1(cons2(X, adx1(L)))
ACTIVE1(incr1(X)) -> INCR1(active1(X))
ACTIVE1(incr1(cons2(X, L))) -> CONS2(s1(X), incr1(L))
ADX1(ok1(X)) -> ADX1(X)
PROPER1(cons2(X1, X2)) -> CONS2(proper1(X1), proper1(X2))
TOP1(ok1(X)) -> TOP1(active1(X))
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 9 SCCs with 22 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TAIL1(mark1(X)) -> TAIL1(X)
TAIL1(ok1(X)) -> TAIL1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
TAIL1(ok1(X)) -> TAIL1(X)
Used argument filtering: TAIL1(x1) = x1
mark1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TAIL1(mark1(X)) -> TAIL1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
TAIL1(mark1(X)) -> TAIL1(X)
Used argument filtering: TAIL1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HEAD1(ok1(X)) -> HEAD1(X)
HEAD1(mark1(X)) -> HEAD1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
HEAD1(mark1(X)) -> HEAD1(X)
Used argument filtering: HEAD1(x1) = x1
ok1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
HEAD1(ok1(X)) -> HEAD1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
HEAD1(ok1(X)) -> HEAD1(X)
Used argument filtering: HEAD1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ADX1(ok1(X)) -> ADX1(X)
ADX1(mark1(X)) -> ADX1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
ADX1(mark1(X)) -> ADX1(X)
Used argument filtering: ADX1(x1) = x1
ok1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ADX1(ok1(X)) -> ADX1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
ADX1(ok1(X)) -> ADX1(X)
Used argument filtering: ADX1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
S1(ok1(X)) -> S1(X)
S1(mark1(X)) -> S1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
S1(mark1(X)) -> S1(X)
Used argument filtering: S1(x1) = x1
ok1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
S1(ok1(X)) -> S1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
S1(ok1(X)) -> S1(X)
Used argument filtering: S1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
CONS2(ok1(X1), ok1(X2)) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2) = x2
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
CONS2(mark1(X1), X2) -> CONS2(X1, X2)
Used argument filtering: CONS2(x1, x2) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INCR1(ok1(X)) -> INCR1(X)
INCR1(mark1(X)) -> INCR1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
INCR1(mark1(X)) -> INCR1(X)
Used argument filtering: INCR1(x1) = x1
ok1(x1) = x1
mark1(x1) = mark1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
INCR1(ok1(X)) -> INCR1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
INCR1(ok1(X)) -> INCR1(X)
Used argument filtering: INCR1(x1) = x1
ok1(x1) = ok1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(tail1(X)) -> PROPER1(X)
PROPER1(head1(X)) -> PROPER1(X)
PROPER1(adx1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
PROPER1(cons2(X1, X2)) -> PROPER1(X1)
PROPER1(cons2(X1, X2)) -> PROPER1(X2)
Used argument filtering: PROPER1(x1) = x1
s1(x1) = x1
cons2(x1, x2) = cons2(x1, x2)
incr1(x1) = x1
tail1(x1) = x1
head1(x1) = x1
adx1(x1) = x1
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(tail1(X)) -> PROPER1(X)
PROPER1(head1(X)) -> PROPER1(X)
PROPER1(adx1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
PROPER1(adx1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
s1(x1) = x1
incr1(x1) = x1
tail1(x1) = x1
head1(x1) = x1
adx1(x1) = adx1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(tail1(X)) -> PROPER1(X)
PROPER1(head1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
PROPER1(head1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
s1(x1) = x1
incr1(x1) = x1
tail1(x1) = x1
head1(x1) = head1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(incr1(X)) -> PROPER1(X)
PROPER1(tail1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
PROPER1(tail1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
s1(x1) = x1
incr1(x1) = x1
tail1(x1) = tail1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(s1(X)) -> PROPER1(X)
PROPER1(incr1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
PROPER1(incr1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
s1(x1) = x1
incr1(x1) = incr1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PROPER1(s1(X)) -> PROPER1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
PROPER1(s1(X)) -> PROPER1(X)
Used argument filtering: PROPER1(x1) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(head1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(tail1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
ACTIVE1(incr1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
ACTIVE1(incr1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
head1(x1) = x1
cons2(x1, x2) = x1
s1(x1) = x1
tail1(x1) = x1
adx1(x1) = x1
incr1(x1) = incr1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(head1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(tail1(X)) -> ACTIVE1(X)
ACTIVE1(adx1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
ACTIVE1(adx1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
head1(x1) = x1
cons2(x1, x2) = x1
s1(x1) = x1
tail1(x1) = x1
adx1(x1) = adx1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(head1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
ACTIVE1(tail1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
ACTIVE1(tail1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
head1(x1) = x1
cons2(x1, x2) = x1
s1(x1) = x1
tail1(x1) = tail1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(head1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
ACTIVE1(s1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
ACTIVE1(s1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
head1(x1) = x1
cons2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(head1(X)) -> ACTIVE1(X)
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
ACTIVE1(cons2(X1, X2)) -> ACTIVE1(X1)
Used argument filtering: ACTIVE1(x1) = x1
head1(x1) = x1
cons2(x1, x2) = cons1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
ACTIVE1(head1(X)) -> ACTIVE1(X)
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.
ACTIVE1(head1(X)) -> ACTIVE1(X)
Used argument filtering: ACTIVE1(x1) = x1
head1(x1) = head1(x1)
Used ordering: Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
TOP1(ok1(X)) -> TOP1(active1(X))
TOP1(mark1(X)) -> TOP1(proper1(X))
The TRS R consists of the following rules:
active1(incr1(nil)) -> mark1(nil)
active1(incr1(cons2(X, L))) -> mark1(cons2(s1(X), incr1(L)))
active1(adx1(nil)) -> mark1(nil)
active1(adx1(cons2(X, L))) -> mark1(incr1(cons2(X, adx1(L))))
active1(nats) -> mark1(adx1(zeros))
active1(zeros) -> mark1(cons2(0, zeros))
active1(head1(cons2(X, L))) -> mark1(X)
active1(tail1(cons2(X, L))) -> mark1(L)
active1(incr1(X)) -> incr1(active1(X))
active1(cons2(X1, X2)) -> cons2(active1(X1), X2)
active1(s1(X)) -> s1(active1(X))
active1(adx1(X)) -> adx1(active1(X))
active1(head1(X)) -> head1(active1(X))
active1(tail1(X)) -> tail1(active1(X))
incr1(mark1(X)) -> mark1(incr1(X))
cons2(mark1(X1), X2) -> mark1(cons2(X1, X2))
s1(mark1(X)) -> mark1(s1(X))
adx1(mark1(X)) -> mark1(adx1(X))
head1(mark1(X)) -> mark1(head1(X))
tail1(mark1(X)) -> mark1(tail1(X))
proper1(incr1(X)) -> incr1(proper1(X))
proper1(nil) -> ok1(nil)
proper1(cons2(X1, X2)) -> cons2(proper1(X1), proper1(X2))
proper1(s1(X)) -> s1(proper1(X))
proper1(adx1(X)) -> adx1(proper1(X))
proper1(nats) -> ok1(nats)
proper1(zeros) -> ok1(zeros)
proper1(0) -> ok1(0)
proper1(head1(X)) -> head1(proper1(X))
proper1(tail1(X)) -> tail1(proper1(X))
incr1(ok1(X)) -> ok1(incr1(X))
cons2(ok1(X1), ok1(X2)) -> ok1(cons2(X1, X2))
s1(ok1(X)) -> ok1(s1(X))
adx1(ok1(X)) -> ok1(adx1(X))
head1(ok1(X)) -> ok1(head1(X))
tail1(ok1(X)) -> ok1(tail1(X))
top1(mark1(X)) -> top1(proper1(X))
top1(ok1(X)) -> top1(active1(X))
Q is empty.
We have to consider all minimal (P,Q,R)-chains.