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

0 QTRS
↳1 QTRSRRRProof (⇔)
↳2 QTRS
↳3 Overlay + Local Confluence (⇔)
↳4 QTRS
↳5 DependencyPairsProof (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 QDP
↳9 UsableRulesProof (⇔)
↳10 QDP
↳11 QReductionProof (⇔)
↳12 QDP
↳13 Instantiation (⇔)
↳14 QDP
↳15 NonTerminationProof (⇔)
↳16 NO

(0) Obligation:

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

f(X) → if(X, c, f(true))
if(true, X, Y) → X
if(false, X, Y) → Y

Q is empty.

(1) QTRSRRRProof (EQUIVALENT transformation)

Used ordering:
Polynomial interpretation [POLO]:

POL(c) = 0   
POL(f(x1)) = x1   
POL(false) = 2   
POL(if(x1, x2, x3)) = x1 + 2·x2 + 2·x3   
POL(true) = 0   
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

if(false, X, Y) → Y


(2) Obligation:

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

f(X) → if(X, c, f(true))
if(true, X, Y) → X

Q is empty.

(3) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(4) Obligation:

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

f(X) → if(X, c, f(true))
if(true, X, Y) → X

The set Q consists of the following terms:

f(x0)
if(true, x0, x1)

(5) DependencyPairsProof (EQUIVALENT transformation)

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

(6) Obligation:

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

F(X) → IF(X, c, f(true))
F(X) → F(true)

The TRS R consists of the following rules:

f(X) → if(X, c, f(true))
if(true, X, Y) → X

The set Q consists of the following terms:

f(x0)
if(true, x0, x1)

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 1 less node.

(8) Obligation:

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

F(X) → F(true)

The TRS R consists of the following rules:

f(X) → if(X, c, f(true))
if(true, X, Y) → X

The set Q consists of the following terms:

f(x0)
if(true, x0, x1)

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:

F(X) → F(true)

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

f(x0)
if(true, x0, x1)

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].

f(x0)
if(true, x0, x1)

(12) Obligation:

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

F(X) → F(true)

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

(13) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule F(X) → F(true) we obtained the following new rules [LPAR04]:

F(true) → F(true)

(14) Obligation:

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

F(true) → F(true)

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

(15) NonTerminationProof (EQUIVALENT 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 = F(true) evaluates to t =F(true)

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 F(true) to F(true).



(16) NO