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:

f1(X) -> if3(X, c, n__f1(n__true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
true -> n__true
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__true) -> true
activate1(X) -> X

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

f1(X) -> if3(X, c, n__f1(n__true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
true -> n__true
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__true) -> true
activate1(X) -> X

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

IF3(false, X, Y) -> ACTIVATE1(Y)
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
F1(X) -> IF3(X, c, n__f1(n__true))
ACTIVATE1(n__true) -> TRUE

The TRS R consists of the following rules:

f1(X) -> if3(X, c, n__f1(n__true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
true -> n__true
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__true) -> true
activate1(X) -> X

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

↳ QTRS
  ↳ DependencyPairsProof
QDP
      ↳ DependencyGraphProof

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

IF3(false, X, Y) -> ACTIVATE1(Y)
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
F1(X) -> IF3(X, c, n__f1(n__true))
ACTIVATE1(n__true) -> TRUE

The TRS R consists of the following rules:

f1(X) -> if3(X, c, n__f1(n__true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
true -> n__true
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__true) -> true
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
QDP
          ↳ QDPOrderProof

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

IF3(false, X, Y) -> ACTIVATE1(Y)
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
F1(X) -> IF3(X, c, n__f1(n__true))

The TRS R consists of the following rules:

f1(X) -> if3(X, c, n__f1(n__true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
true -> n__true
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__true) -> true
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__f1(X)) -> ACTIVATE1(X)
The remaining pairs can at least be oriented weakly.

IF3(false, X, Y) -> ACTIVATE1(Y)
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
F1(X) -> IF3(X, c, n__f1(n__true))
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE1(x1)) = x1   
POL(F1(x1)) = 1   
POL(IF3(x1, x2, x3)) = x3   
POL(activate1(x1)) = 0   
POL(c) = 0   
POL(f1(x1)) = 0   
POL(false) = 0   
POL(if3(x1, x2, x3)) = 0   
POL(n__f1(x1)) = 1 + x1   
POL(n__true) = 0   
POL(true) = 0   

The following usable rules [14] were oriented: none



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

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

IF3(false, X, Y) -> ACTIVATE1(Y)
ACTIVATE1(n__f1(X)) -> F1(activate1(X))
F1(X) -> IF3(X, c, n__f1(n__true))

The TRS R consists of the following rules:

f1(X) -> if3(X, c, n__f1(n__true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
true -> n__true
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__true) -> true
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be oriented strictly and are deleted.


ACTIVATE1(n__f1(X)) -> F1(activate1(X))
The remaining pairs can at least be oriented weakly.

IF3(false, X, Y) -> ACTIVATE1(Y)
F1(X) -> IF3(X, c, n__f1(n__true))
Used ordering: Polynomial interpretation [21]:

POL(ACTIVATE1(x1)) = 1 + x1   
POL(F1(x1)) = x1   
POL(IF3(x1, x2, x3)) = x1 + x3   
POL(activate1(x1)) = x1   
POL(c) = 0   
POL(f1(x1)) = x1   
POL(false) = 1   
POL(if3(x1, x2, x3)) = x2 + x3   
POL(n__f1(x1)) = x1   
POL(n__true) = 0   
POL(true) = 0   

The following usable rules [14] were oriented:

if3(false, X, Y) -> activate1(Y)
f1(X) -> if3(X, c, n__f1(n__true))
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__true) -> true
true -> n__true
activate1(X) -> X
f1(X) -> n__f1(X)
if3(true, X, Y) -> X



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

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

IF3(false, X, Y) -> ACTIVATE1(Y)
F1(X) -> IF3(X, c, n__f1(n__true))

The TRS R consists of the following rules:

f1(X) -> if3(X, c, n__f1(n__true))
if3(true, X, Y) -> X
if3(false, X, Y) -> activate1(Y)
f1(X) -> n__f1(X)
true -> n__true
activate1(n__f1(X)) -> f1(activate1(X))
activate1(n__true) -> true
activate1(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 2 less nodes.