Termination w.r.t. Q of the following Term Rewriting System could be disproven:
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
and(tt, X) → activate(X)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
and(tt, X) → activate(X)
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(activate(x1)) = x1
POL(and(x1, x2)) = 1 + 2·x1 + 2·x2
POL(cons(x1, x2)) = 2·x1 + 2·x2
POL(length(x1)) = 2·x1
POL(n__zeros) = 0
POL(nil) = 0
POL(s(x1)) = 2·x1
POL(tt) = 0
POL(zeros) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(nil) → 0
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
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(activate(x1)) = x1
POL(cons(x1, x2)) = 2·x1 + x2
POL(length(x1)) = 2·x1
POL(n__zeros) = 0
POL(nil) = 2
POL(s(x1)) = x1
POL(zeros) = 0
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
zeros → n__zeros
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
zeros → n__zeros
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(activate(x1)) = 2·x1
POL(cons(x1, x2)) = x1 + 2·x2
POL(length(x1)) = 2·x1
POL(n__zeros) = 1
POL(s(x1)) = x1
POL(zeros) = 2
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
activate(X) → X
Q is empty.
The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:
activate(X) → X
Used ordering:
Polynomial interpretation [25]:
POL(0) = 0
POL(activate(x1)) = 1 + x1
POL(cons(x1, x2)) = 1 + 2·x1 + x2
POL(length(x1)) = 2·x1
POL(n__zeros) = 0
POL(s(x1)) = x1
POL(zeros) = 1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
Q is empty.
The TRS is overlay and locally confluent. By [19] we can switch to innermost.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
activate(n__zeros)
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
LENGTH(cons(N, L)) → ACTIVATE(L)
ACTIVATE(n__zeros) → ZEROS
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
activate(n__zeros)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
LENGTH(cons(N, L)) → ACTIVATE(L)
ACTIVATE(n__zeros) → ZEROS
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
activate(n__zeros)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
length(cons(N, L)) → s(length(activate(L)))
activate(n__zeros) → zeros
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
activate(n__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
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
length(cons(x0, x1))
activate(n__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.
length(cons(x0, x1))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(N, L)) → LENGTH(activate(L))
The TRS R consists of the following rules:
activate(n__zeros) → zeros
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
activate(n__zeros)
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LENGTH(cons(N, L)) → LENGTH(activate(L)) at position [0] we obtained the following new rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
The TRS R consists of the following rules:
activate(n__zeros) → zeros
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
activate(n__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
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
activate(n__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(n__zeros)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(zeros)
The TRS R consists of the following rules:
zeros → cons(0, n__zeros)
The set Q consists of the following terms:
zeros
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule LENGTH(cons(y0, n__zeros)) → LENGTH(zeros) at position [0] we obtained the following new rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
The TRS R consists of the following rules:
zeros → cons(0, 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
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(cons(0, n__zeros))
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
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Instantiation
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(y0, n__zeros)) → LENGTH(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)) → LENGTH(cons(0, n__zeros)) we obtained the following new rules:
LENGTH(cons(0, n__zeros)) → LENGTH(cons(0, n__zeros))
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ Overlay + Local Confluence
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Instantiation
↳ QDP
↳ NonTerminationProof
Q DP problem:
The TRS P consists of the following rules:
LENGTH(cons(0, n__zeros)) → LENGTH(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 semiunifying a rule from P directly.
The TRS P consists of the following rules:
LENGTH(cons(0, n__zeros)) → LENGTH(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
The DP semiunifies directly so there is only one rewrite step from LENGTH(cons(0, n__zeros)) to LENGTH(cons(0, n__zeros)).