Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

Q is empty.

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, n__f(true))
ACTIVATE(n__f(X)) → F(X)
IF(false, X, Y) → ACTIVATE(Y)

The TRS R consists of the following rules:

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

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof

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

F(X) → IF(X, c, n__f(true))
ACTIVATE(n__f(X)) → F(X)
IF(false, X, Y) → ACTIVATE(Y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(X) → IF(X, c, n__f(true))
ACTIVATE(n__f(X)) → F(X)
The remaining pairs can at least be oriented weakly.

IF(false, X, Y) → ACTIVATE(Y)
Used ordering: Polynomial interpretation [25,35]:

POL(c) = 1/4   
POL(n__f(x1)) = 3/4 + (2)x_1   
POL(true) = 0   
POL(false) = 4   
POL(IF(x1, x2, x3)) = (1/2)x_1 + (3/4)x_3   
POL(ACTIVATE(x1)) = 2 + (3/4)x_1   
POL(F(x1)) = 2 + (3/2)x_1   
The value of delta used in the strict ordering is 9/16.
The following usable rules [17] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
        ↳ QDP
          ↳ DependencyGraphProof

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

IF(false, X, Y) → ACTIVATE(Y)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.