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:
zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
The set Q consists of the following terms:
activate(x0)
length(cons(x0, x1))
zeros
length(nil)
U12(tt, x0)
U11(tt, x0)
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
The set Q consists of the following terms:
activate(x0)
length(cons(x0, x1))
zeros
length(nil)
U12(tt, x0)
U11(tt, x0)
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(nil) → 0
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
The set Q consists of the following terms:
activate(x0)
length(cons(x0, x1))
zeros
length(nil)
U12(tt, x0)
U11(tt, x0)
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
length(nil) → 0
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(U11(x1, x2)) = 2·x1 + 2·x2
POL(U12(x1, x2)) = 2·x1 + 2·x2
POL(activate(x1)) = x1
POL(cons(x1, x2)) = 2·x1 + 2·x2
POL(length(x1)) = 2·x1
POL(n__zeros) = 0
POL(nil) = 1
POL(s(x1)) = x1
POL(tt) = 0
POL(zeros) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
The set Q consists of the following terms:
activate(x0)
length(cons(x0, x1))
zeros
length(nil)
U12(tt, x0)
U11(tt, x0)
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
U121(tt, L) → ACTIVATE(L)
LENGTH(cons(N, L)) → ACTIVATE(L)
U121(tt, L) → LENGTH(activate(L))
U111(tt, L) → U121(tt, activate(L))
U111(tt, L) → ACTIVATE(L)
LENGTH(cons(N, L)) → U111(tt, activate(L))
ACTIVATE(n__zeros) → ZEROS
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
The set Q consists of the following terms:
activate(x0)
length(cons(x0, x1))
zeros
length(nil)
U12(tt, x0)
U11(tt, x0)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, L) → ACTIVATE(L)
LENGTH(cons(N, L)) → ACTIVATE(L)
U121(tt, L) → LENGTH(activate(L))
U111(tt, L) → U121(tt, activate(L))
U111(tt, L) → ACTIVATE(L)
LENGTH(cons(N, L)) → U111(tt, activate(L))
ACTIVATE(n__zeros) → ZEROS
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
The set Q consists of the following terms:
activate(x0)
length(cons(x0, x1))
zeros
length(nil)
U12(tt, x0)
U11(tt, x0)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, L) → LENGTH(activate(L))
U111(tt, L) → U121(tt, activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
U11(tt, L) → U12(tt, activate(L))
U12(tt, L) → s(length(activate(L)))
length(cons(N, L)) → U11(tt, activate(L))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
The set Q consists of the following terms:
activate(x0)
length(cons(x0, x1))
zeros
length(nil)
U12(tt, x0)
U11(tt, x0)
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, L) → LENGTH(activate(L))
U111(tt, L) → U121(tt, activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
activate(x0)
length(cons(x0, x1))
zeros
length(nil)
U12(tt, x0)
U11(tt, x0)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
length(cons(x0, x1))
length(nil)
U12(tt, x0)
U11(tt, x0)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, L) → LENGTH(activate(L))
U111(tt, L) → U121(tt, activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
activate(x0)
zeros
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U121(tt, L) → LENGTH(activate(L)) at position [0] we obtained the following new rules:
U121(tt, n__zeros) → LENGTH(zeros)
U121(tt, x0) → LENGTH(x0)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U111(tt, L) → U121(tt, activate(L))
U121(tt, n__zeros) → LENGTH(zeros)
LENGTH(cons(N, L)) → U111(tt, activate(L))
U121(tt, x0) → LENGTH(x0)
The TRS R consists of the following rules:
activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
activate(x0)
zeros
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U111(tt, L) → U121(tt, activate(L)) at position [1] we obtained the following new rules:
U111(tt, x0) → U121(tt, x0)
U111(tt, n__zeros) → U121(tt, zeros)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U111(tt, n__zeros) → U121(tt, zeros)
U121(tt, n__zeros) → LENGTH(zeros)
LENGTH(cons(N, L)) → U111(tt, activate(L))
U111(tt, x0) → U121(tt, x0)
U121(tt, x0) → LENGTH(x0)
The TRS R consists of the following rules:
activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
activate(x0)
zeros
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LENGTH(cons(N, L)) → U111(tt, activate(L)) at position [1] we obtained the following new rules:
LENGTH(cons(y0, x0)) → U111(tt, x0)
LENGTH(cons(y0, n__zeros)) → U111(tt, zeros)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → U111(tt, zeros)
U111(tt, n__zeros) → U121(tt, zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U121(tt, n__zeros) → LENGTH(zeros)
U121(tt, x0) → LENGTH(x0)
U111(tt, x0) → U121(tt, x0)
The TRS R consists of the following rules:
activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
activate(x0)
zeros
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → U111(tt, zeros)
U111(tt, n__zeros) → U121(tt, zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U121(tt, n__zeros) → LENGTH(zeros)
U111(tt, x0) → U121(tt, x0)
U121(tt, x0) → LENGTH(x0)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
activate(x0)
zeros
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
activate(x0)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → U111(tt, zeros)
U111(tt, n__zeros) → U121(tt, zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U121(tt, n__zeros) → LENGTH(zeros)
U121(tt, x0) → LENGTH(x0)
U111(tt, x0) → U121(tt, x0)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LENGTH(cons(y0, n__zeros)) → U111(tt, zeros) at position [1] we obtained the following new rules:
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U111(tt, n__zeros) → U121(tt, zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U121(tt, n__zeros) → LENGTH(zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
U111(tt, x0) → U121(tt, x0)
U121(tt, x0) → LENGTH(x0)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U111(tt, n__zeros) → U121(tt, zeros) at position [1] we obtained the following new rules:
U111(tt, n__zeros) → U121(tt, n__zeros)
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U111(tt, n__zeros) → U121(tt, n__zeros)
U121(tt, n__zeros) → LENGTH(zeros)
U121(tt, x0) → LENGTH(x0)
U111(tt, x0) → U121(tt, x0)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule U121(tt, n__zeros) → LENGTH(zeros) at position [0] we obtained the following new rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, n__zeros) → LENGTH(n__zeros)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, n__zeros) → LENGTH(n__zeros)
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
U111(tt, x0) → U121(tt, x0)
U121(tt, x0) → LENGTH(x0)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
U121(tt, x0) → LENGTH(x0)
U111(tt, x0) → U121(tt, x0)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
U121(tt, x0) → LENGTH(x0)
U111(tt, x0) → U121(tt, x0)
R is empty.
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
zeros
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U111(tt, n__zeros) → U121(tt, n__zeros)
U111(tt, x0) → U121(tt, x0)
U121(tt, x0) → LENGTH(x0)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule U121(tt, x0) → LENGTH(x0) we obtained the following new rules:
U121(tt, cons(y_0, n__zeros)) → LENGTH(cons(y_0, n__zeros))
U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(y_0, n__zeros)) → LENGTH(cons(y_0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
LENGTH(cons(y0, x0)) → U111(tt, x0)
U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1))
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
U111(tt, x0) → U121(tt, x0)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule U111(tt, x0) → U121(tt, x0) we obtained the following new rules:
U111(tt, cons(y_0, y_1)) → U121(tt, cons(y_0, y_1))
U111(tt, cons(y_0, n__zeros)) → U121(tt, cons(y_0, n__zeros))
U111(tt, n__zeros) → U121(tt, n__zeros)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(y_0, n__zeros)) → LENGTH(cons(y_0, n__zeros))
U111(tt, cons(y_0, y_1)) → U121(tt, cons(y_0, y_1))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
U111(tt, cons(y_0, n__zeros)) → U121(tt, cons(y_0, n__zeros))
LENGTH(cons(y0, x0)) → U111(tt, x0)
U111(tt, n__zeros) → U121(tt, n__zeros)
U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1))
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule LENGTH(cons(y0, x0)) → U111(tt, x0) we obtained the following new rules:
LENGTH(cons(x0, cons(y_0, y_1))) → U111(tt, cons(y_0, y_1))
LENGTH(cons(x0, cons(y_0, n__zeros))) → U111(tt, cons(y_0, n__zeros))
LENGTH(cons(x0, n__zeros)) → U111(tt, n__zeros)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(x0, cons(y_0, y_1))) → U111(tt, cons(y_0, y_1))
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(y_0, n__zeros)) → LENGTH(cons(y_0, n__zeros))
U111(tt, cons(y_0, y_1)) → U121(tt, cons(y_0, y_1))
LENGTH(cons(x0, cons(y_0, n__zeros))) → U111(tt, cons(y_0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
U111(tt, cons(y_0, n__zeros)) → U121(tt, cons(y_0, n__zeros))
U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1))
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
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.
LENGTH(cons(x0, cons(y_0, y_1))) → U111(tt, cons(y_0, y_1))
LENGTH(cons(x0, cons(y_0, n__zeros))) → U111(tt, cons(y_0, n__zeros))
The remaining pairs can at least be oriented weakly.
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(y_0, n__zeros)) → LENGTH(cons(y_0, n__zeros))
U111(tt, cons(y_0, y_1)) → U121(tt, cons(y_0, y_1))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
U111(tt, cons(y_0, n__zeros)) → U121(tt, cons(y_0, n__zeros))
U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1))
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
Used ordering: Matrix interpretation [3]:
Non-tuple symbols:
| M( cons(x1, x2) ) = | | + | | · | x1 | + | | · | x2 |
Tuple symbols:
| M( U121(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
| M( U111(x1, x2) ) = | 0 | + | | · | x1 | + | | · | x2 |
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:
none
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Instantiation
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(y_0, n__zeros)) → LENGTH(cons(y_0, n__zeros))
U111(tt, cons(y_0, y_1)) → U121(tt, cons(y_0, y_1))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
U111(tt, cons(y_0, n__zeros)) → U121(tt, cons(y_0, n__zeros))
U111(tt, n__zeros) → U121(tt, n__zeros)
U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1))
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule U111(tt, cons(y_0, y_1)) → U121(tt, cons(y_0, y_1)) we obtained the following new rules:
U111(tt, cons(0, n__zeros)) → U121(tt, cons(0, n__zeros))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(y_0, n__zeros)) → LENGTH(cons(y_0, n__zeros))
U111(tt, cons(0, n__zeros)) → U121(tt, cons(0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
U111(tt, cons(y_0, n__zeros)) → U121(tt, cons(y_0, n__zeros))
U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1))
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule U111(tt, cons(y_0, n__zeros)) → U121(tt, cons(y_0, n__zeros)) we obtained the following new rules:
U111(tt, cons(0, n__zeros)) → U121(tt, cons(0, n__zeros))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(y_0, n__zeros)) → LENGTH(cons(y_0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
U111(tt, cons(0, n__zeros)) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
U111(tt, n__zeros) → U121(tt, n__zeros)
U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1))
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule U121(tt, cons(y_0, n__zeros)) → LENGTH(cons(y_0, n__zeros)) we obtained the following new rules:
U121(tt, cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U111(tt, cons(0, n__zeros)) → U121(tt, cons(0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1))
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule U121(tt, cons(y_0, y_1)) → LENGTH(cons(y_0, y_1)) we obtained the following new rules:
U121(tt, cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
U111(tt, cons(0, n__zeros)) → U121(tt, cons(0, n__zeros))
LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros)
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule LENGTH(cons(y0, n__zeros)) → U111(tt, n__zeros) we obtained the following new rules:
LENGTH(cons(0, n__zeros)) → U111(tt, n__zeros)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U111(tt, cons(0, n__zeros)) → U121(tt, cons(0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
LENGTH(cons(0, n__zeros)) → U111(tt, n__zeros)
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule LENGTH(cons(y0, n__zeros)) → U111(tt, cons(0, n__zeros)) we obtained the following new rules:
LENGTH(cons(0, n__zeros)) → U111(tt, cons(0, n__zeros))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ Narrowing
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ ForwardInstantiation
↳ QDP
↳ QDPOrderProof
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ Instantiation
↳ QDP
↳ NonTerminationProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
U111(tt, cons(0, n__zeros)) → U121(tt, cons(0, n__zeros))
LENGTH(cons(0, n__zeros)) → U111(tt, n__zeros)
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(0, n__zeros)) → U111(tt, cons(0, n__zeros))
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We used the non-termination processor [17] to show that the DP problem is infinite.
Found a loop by narrowing to the left:
The TRS P consists of the following rules:
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
U121(tt, cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
U111(tt, n__zeros) → U121(tt, cons(0, n__zeros))
U111(tt, cons(0, n__zeros)) → U121(tt, cons(0, n__zeros))
LENGTH(cons(0, n__zeros)) → U111(tt, n__zeros)
U111(tt, n__zeros) → U121(tt, n__zeros)
LENGTH(cons(0, n__zeros)) → U111(tt, cons(0, n__zeros))
The TRS R consists of the following rules:none
s = LENGTH(cons(0, n__zeros)) evaluates to t =LENGTH(cons(0, n__zeros))
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
LENGTH(cons(0, n__zeros)) → U111(tt, n__zeros)
with rule LENGTH(cons(0, n__zeros)) → U111(tt, n__zeros) at position [] and matcher [ ]
U111(tt, n__zeros) → U121(tt, n__zeros)
with rule U111(tt, n__zeros) → U121(tt, n__zeros) at position [] and matcher [ ]
U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
with rule U121(tt, n__zeros) → LENGTH(cons(0, n__zeros))
Now applying the matcher to the start term leads to a term which is equal to the last term in the rewriting sequence
All these steps are and every following step will be a correct step w.r.t to Q.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
U121(tt, L) → LENGTH(activate(L))
U111(tt, L) → U121(tt, activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
activate(x0)
length(cons(x0, x1))
zeros
length(nil)
U12(tt, x0)
U11(tt, x0)
We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.
length(cons(x0, x1))
length(nil)
U12(tt, x0)
U11(tt, x0)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
U121(tt, L) → LENGTH(activate(L))
U111(tt, L) → U121(tt, activate(L))
LENGTH(cons(N, L)) → U111(tt, activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
activate(X) → X
zeros → cons(0, n__zeros)
zeros → n__zeros
The set Q consists of the following terms:
activate(x0)
zeros
We have to consider all minimal (P,Q,R)-chains.