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:
g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
c → n__c
activate(n__d) → d
activate(n__c) → c
activate(X) → X
Q is empty.
↳ QTRS
↳ RRRPoloQTRSProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
c → n__c
activate(n__d) → d
activate(n__c) → c
activate(X) → X
Q is empty.
The following Q TRS is given: Q restricted rewrite system:
The TRS R consists of the following rules:
g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
c → n__c
activate(n__d) → d
activate(n__c) → c
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:
c → n__c
activate(n__d) → d
activate(X) → X
Used ordering:
Polynomial interpretation [25]:
POL(activate(x1)) = 1 + 2·x1
POL(c) = 1
POL(d) = 1
POL(g(x1)) = 1 + 2·x1
POL(h(x1)) = x1
POL(n__c) = 0
POL(n__d) = 1
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
Q restricted rewrite system:
The TRS R consists of the following rules:
g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c
Q is empty.
We have applied [19,8] to switch to innermost. The TRS R 1 is
d → n__d
activate(n__c) → c
c → d
The TRS R 2 is
g(X) → h(activate(X))
h(n__d) → g(n__c)
The signature Sigma is {g, h}
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
Q restricted rewrite system:
The TRS R consists of the following rules:
g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c
The set Q consists of the following terms:
g(x0)
c
h(n__d)
d
activate(n__c)
Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
C → D
ACTIVATE(n__c) → C
G(X) → H(activate(X))
G(X) → ACTIVATE(X)
The TRS R consists of the following rules:
g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c
The set Q consists of the following terms:
g(x0)
c
h(n__d)
d
activate(n__c)
We have to consider all minimal (P,Q,R)-chains.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
C → D
ACTIVATE(n__c) → C
G(X) → H(activate(X))
G(X) → ACTIVATE(X)
The TRS R consists of the following rules:
g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c
The set Q consists of the following terms:
g(x0)
c
h(n__d)
d
activate(n__c)
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 3 less nodes.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(X) → H(activate(X))
The TRS R consists of the following rules:
g(X) → h(activate(X))
c → d
h(n__d) → g(n__c)
d → n__d
activate(n__c) → c
The set Q consists of the following terms:
g(x0)
c
h(n__d)
d
activate(n__c)
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(X) → H(activate(X))
The TRS R consists of the following rules:
activate(n__c) → c
c → d
d → n__d
The set Q consists of the following terms:
g(x0)
c
h(n__d)
d
activate(n__c)
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.
g(x0)
h(n__d)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(X) → H(activate(X))
The TRS R consists of the following rules:
activate(n__c) → c
c → d
d → n__d
The set Q consists of the following terms:
c
d
activate(n__c)
We have to consider all minimal (P,Q,R)-chains.
We use the modular non-overlap check [17] to decrease Q to the empty set.
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ MNOCProof
↳ QDP
↳ NonTerminationProof
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(X) → H(activate(X))
The TRS R consists of the following rules:
activate(n__c) → c
c → d
d → n__d
Q is empty.
We have to consider all (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:
H(n__d) → G(n__c)
G(X) → H(activate(X))
The TRS R consists of the following rules:
activate(n__c) → c
c → d
d → n__d
s = G(n__c) evaluates to t =G(n__c)
Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
- Semiunifier: [ ]
- Matcher: [ ]
Rewriting sequence
G(n__c) → H(activate(n__c))
with rule G(X) → H(activate(X)) at position [] and matcher [X / n__c]
H(activate(n__c)) → H(c)
with rule activate(n__c) → c at position [0] and matcher [ ]
H(c) → H(d)
with rule c → d at position [0] and matcher [ ]
H(d) → H(n__d)
with rule d → n__d at position [0] and matcher [ ]
H(n__d) → G(n__c)
with rule H(n__d) → G(n__c)
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(X) → H(activate(X))
The TRS R consists of the following rules:
activate(n__c) → c
c → d
d → n__d
The set Q consists of the following terms:
g(x0)
c
h(n__d)
d
activate(n__c)
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.
g(x0)
h(n__d)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(X) → H(activate(X))
The TRS R consists of the following rules:
activate(n__c) → c
c → d
d → n__d
The set Q consists of the following terms:
c
d
activate(n__c)
We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule G(X) → H(activate(X)) at position [0] we obtained the following new rules:
G(n__c) → H(c)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
G(n__c) → H(c)
H(n__d) → G(n__c)
The TRS R consists of the following rules:
activate(n__c) → c
c → d
d → n__d
The set Q consists of the following terms:
c
d
activate(n__c)
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
G(n__c) → H(c)
H(n__d) → G(n__c)
The TRS R consists of the following rules:
c → d
d → n__d
The set Q consists of the following terms:
c
d
activate(n__c)
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__c)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
G(n__c) → H(c)
H(n__d) → G(n__c)
The TRS R consists of the following rules:
c → d
d → n__d
The set Q consists of the following terms:
c
d
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule G(n__c) → H(c) at position [0] we obtained the following new rules:
G(n__c) → H(d)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(n__c) → H(d)
The TRS R consists of the following rules:
c → d
d → n__d
The set Q consists of the following terms:
c
d
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ 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:
H(n__d) → G(n__c)
G(n__c) → H(d)
The TRS R consists of the following rules:
d → n__d
The set Q consists of the following terms:
c
d
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.
c
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(n__c) → H(d)
The TRS R consists of the following rules:
d → n__d
The set Q consists of the following terms:
d
We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule G(n__c) → H(d) at position [0] we obtained the following new rules:
G(n__c) → H(n__d)
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(n__c) → H(n__d)
The TRS R consists of the following rules:
d → n__d
The set Q consists of the following terms:
d
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
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(n__c) → H(n__d)
R is empty.
The set Q consists of the following terms:
d
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.
d
↳ QTRS
↳ RRRPoloQTRSProof
↳ QTRS
↳ AAECC Innermost
↳ QTRS
↳ DependencyPairsProof
↳ QDP
↳ DependencyGraphProof
↳ QDP
↳ UsableRulesProof
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Narrowing
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
↳ Rewriting
↳ QDP
↳ UsableRulesProof
↳ QDP
↳ QReductionProof
↳ QDP
Q DP problem:
The TRS P consists of the following rules:
H(n__d) → G(n__c)
G(n__c) → H(n__d)
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.