Termination w.r.t. Q of the following Term Rewriting System could be proven:
Q restricted rewrite system:
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> X
Q is empty.
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> X
Q is empty.
Q DP problem:
The TRS P consists of the following rules:
2NDSNEG2(s1(N), cons22(X, cons2(Y, Z))) -> 2NDSPOS2(N, activate1(Z))
2NDSPOS2(s1(N), cons22(X, cons2(Y, Z))) -> ACTIVATE1(Z)
2NDSNEG2(s1(N), cons22(X, cons2(Y, Z))) -> ACTIVATE1(Z)
ACTIVATE1(n__from1(X)) -> FROM1(X)
SQUARE1(X) -> TIMES2(X, X)
PI1(X) -> FROM1(0)
2NDSPOS2(s1(N), cons2(X, Z)) -> 2NDSPOS2(s1(N), cons22(X, activate1(Z)))
PI1(X) -> 2NDSPOS2(X, from1(0))
PLUS2(s1(X), Y) -> PLUS2(X, Y)
TIMES2(s1(X), Y) -> PLUS2(Y, times2(X, Y))
TIMES2(s1(X), Y) -> TIMES2(X, Y)
2NDSPOS2(s1(N), cons22(X, cons2(Y, Z))) -> 2NDSNEG2(N, activate1(Z))
2NDSNEG2(s1(N), cons2(X, Z)) -> 2NDSNEG2(s1(N), cons22(X, activate1(Z)))
2NDSNEG2(s1(N), cons2(X, Z)) -> ACTIVATE1(Z)
2NDSPOS2(s1(N), cons2(X, Z)) -> ACTIVATE1(Z)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> 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:
2NDSNEG2(s1(N), cons22(X, cons2(Y, Z))) -> 2NDSPOS2(N, activate1(Z))
2NDSPOS2(s1(N), cons22(X, cons2(Y, Z))) -> ACTIVATE1(Z)
2NDSNEG2(s1(N), cons22(X, cons2(Y, Z))) -> ACTIVATE1(Z)
ACTIVATE1(n__from1(X)) -> FROM1(X)
SQUARE1(X) -> TIMES2(X, X)
PI1(X) -> FROM1(0)
2NDSPOS2(s1(N), cons2(X, Z)) -> 2NDSPOS2(s1(N), cons22(X, activate1(Z)))
PI1(X) -> 2NDSPOS2(X, from1(0))
PLUS2(s1(X), Y) -> PLUS2(X, Y)
TIMES2(s1(X), Y) -> PLUS2(Y, times2(X, Y))
TIMES2(s1(X), Y) -> TIMES2(X, Y)
2NDSPOS2(s1(N), cons22(X, cons2(Y, Z))) -> 2NDSNEG2(N, activate1(Z))
2NDSNEG2(s1(N), cons2(X, Z)) -> 2NDSNEG2(s1(N), cons22(X, activate1(Z)))
2NDSNEG2(s1(N), cons2(X, Z)) -> ACTIVATE1(Z)
2NDSPOS2(s1(N), cons2(X, Z)) -> ACTIVATE1(Z)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 3 SCCs with 9 less nodes.
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
PLUS2(s1(X), Y) -> PLUS2(X, Y)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> 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.
PLUS2(s1(X), Y) -> PLUS2(X, Y)
Used argument filtering: PLUS2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> 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
Q DP problem:
The TRS P consists of the following rules:
TIMES2(s1(X), Y) -> TIMES2(X, Y)
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> 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.
TIMES2(s1(X), Y) -> TIMES2(X, Y)
Used argument filtering: TIMES2(x1, x2) = x1
s1(x1) = s1(x1)
Used ordering: Quasi Precedence:
trivial
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ PisEmptyProof
↳ QDP
Q DP problem:
P is empty.
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> 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
Q DP problem:
The TRS P consists of the following rules:
2NDSNEG2(s1(N), cons22(X, cons2(Y, Z))) -> 2NDSPOS2(N, activate1(Z))
2NDSNEG2(s1(N), cons2(X, Z)) -> 2NDSNEG2(s1(N), cons22(X, activate1(Z)))
2NDSPOS2(s1(N), cons22(X, cons2(Y, Z))) -> 2NDSNEG2(N, activate1(Z))
2NDSPOS2(s1(N), cons2(X, Z)) -> 2NDSPOS2(s1(N), cons22(X, activate1(Z)))
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> 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.
2NDSNEG2(s1(N), cons22(X, cons2(Y, Z))) -> 2NDSPOS2(N, activate1(Z))
2NDSPOS2(s1(N), cons22(X, cons2(Y, Z))) -> 2NDSNEG2(N, activate1(Z))
Used argument filtering: 2NDSNEG2(x1, x2) = x1
s1(x1) = s1(x1)
2NDSPOS2(x1, x2) = x1
activate1(x1) = x1
n__from1(x1) = n__from
from1(x1) = from
cons2(x1, x2) = cons
Used ordering: Quasi Precedence:
[n__from, from, cons]
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ AND
↳ QDP
↳ QDP
↳ QDP
↳ QDPAfsSolverProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
2NDSNEG2(s1(N), cons2(X, Z)) -> 2NDSNEG2(s1(N), cons22(X, activate1(Z)))
2NDSPOS2(s1(N), cons2(X, Z)) -> 2NDSPOS2(s1(N), cons22(X, activate1(Z)))
The TRS R consists of the following rules:
from1(X) -> cons2(X, n__from1(s1(X)))
2ndspos2(0, Z) -> rnil
2ndspos2(s1(N), cons2(X, Z)) -> 2ndspos2(s1(N), cons22(X, activate1(Z)))
2ndspos2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(posrecip1(Y), 2ndsneg2(N, activate1(Z)))
2ndsneg2(0, Z) -> rnil
2ndsneg2(s1(N), cons2(X, Z)) -> 2ndsneg2(s1(N), cons22(X, activate1(Z)))
2ndsneg2(s1(N), cons22(X, cons2(Y, Z))) -> rcons2(negrecip1(Y), 2ndspos2(N, activate1(Z)))
pi1(X) -> 2ndspos2(X, from1(0))
plus2(0, Y) -> Y
plus2(s1(X), Y) -> s1(plus2(X, Y))
times2(0, Y) -> 0
times2(s1(X), Y) -> plus2(Y, times2(X, Y))
square1(X) -> times2(X, X)
from1(X) -> n__from1(X)
activate1(n__from1(X)) -> from1(X)
activate1(X) -> X
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 0 SCCs with 2 less nodes.