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:

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

The set Q consists of the following terms:

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



↳ QTRS
  ↳ RRRPoloQTRSProof

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

The set Q consists of the following terms:

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


The following Q TRS is given: 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

The set Q consists of the following terms:

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

The following rules can be removed by the rule removal processor [15] because they are oriented strictly by a polynomial ordering:

if(false, X, Y) → Y
Used ordering:
Polynomial interpretation [25]:

POL(c) = 0   
POL(f(x1)) = x1   
POL(false) = 2   
POL(if(x1, x2, x3)) = x1 + 2·x2 + 2·x3   
POL(true) = 0   




↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof

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:

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


Using Dependency Pairs [1,15] we result in the following initial DP problem:
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:

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

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:

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:

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

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

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:

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

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

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:

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

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.

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



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Instantiation

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.
By instantiating [15] the rule F(X) → F(true) we obtained the following new rules:

F(true) → F(true)



↳ QTRS
  ↳ RRRPoloQTRSProof
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ UsableRulesProof
                ↳ QDP
                  ↳ QReductionProof
                    ↳ QDP
                      ↳ Instantiation
                        ↳ QDP
                          ↳ NonTerminationProof

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.
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:

F(true) → F(true)

The TRS R consists of the following rules:none


s = F(true) evaluates to t =F(true)

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:



Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from F(true) to F(true).