NO Termination w.r.t. Q proof of Transformed_CSR_04_Ex1_GL02a_L.ari

Termination w.r.t. Q of the given QTRS could be disproven:

0 QTRS
↳1 QTRSRRRProof (⇔, 0 ms)
↳2 QTRS
↳3 QTRSRRRProof (⇔, 0 ms)
↳4 QTRS
↳5 AAECC Innermost (⇔, 0 ms)
↳6 QTRS
↳7 DependencyPairsProof (⇔, 8 ms)
↳8 QDP
↳9 UsableRulesProof (⇔, 0 ms)
↳10 QDP
↳11 QReductionProof (⇔, 0 ms)
↳12 QDP
↳13 NonTerminationLoopProof (⇒, 0 ms)
↳14 NO

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

eq → true
eq → eq
eq → false
inf(X) → cons
take(0, X) → nil
take(s, cons) → cons
length(nil) → 0
length(cons) → s

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(0) = 0   
POL(cons) = 0   
POL(eq) = 2   
POL(false) = 0   
POL(inf(x1)) = x1   
POL(length(x1)) = x1   
POL(nil) = 0   
POL(s) = 0   
POL(take(x1, x2)) = 2·x1 + x2   
POL(true) = 1   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

eq → true
eq → false


(2) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

eq → eq
inf(X) → cons
take(0, X) → nil
take(s, cons) → cons
length(nil) → 0
length(cons) → s

Q is empty.

(3) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Knuth-Bendix order [KBO] with precedence:
length1 > s > nil > 0 > take2 > inf1 > cons > eq

and weight map:

eq=1
cons=2
0=2
nil=1
s=3
inf_1=1
length_1=1
take_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

inf(X) → cons
take(0, X) → nil
take(s, cons) → cons
length(nil) → 0
length(cons) → s


(4) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

eq → eq

Q is empty.

(5) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is

eq → eq

The signature Sigma is {eq}

(6) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

eq → eq

The set Q consists of the following terms:

eq

(7) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(8) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ → EQ

The TRS R consists of the following rules:

eq → eq

The set Q consists of the following terms:

eq

We have to consider all minimal (P,Q,R)-chains.

(9) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ → EQ

R is empty.
The set Q consists of the following terms:

eq

We have to consider all minimal (P,Q,R)-chains.

(11) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

eq

(12) Obligation:

Q DP problem:
The TRS P consists of the following rules:

EQ → EQ

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(13) NonTerminationLoopProof (COMPLETE transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = EQ evaluates to t =EQ

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:
  • Matcher: [ ]
  • Semiunifier: [ ]



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from EQ to EQ.



(14) NO