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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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


F1(X) -> IF3(X, c, n__f1(true))
The remaining pairs can at least be oriented weakly.

ACTIVATE1(n__f1(X)) -> F1(X)
IF3(false, X, Y) -> ACTIVATE1(Y)
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( true ) = max{0, -3}


POL( false ) = 3


POL( ACTIVATE1(x1) ) = x1


POL( IF3(x1, ..., x3) ) = max{0, x1 + 2x3 - 3}


POL( c ) = max{0, -3}


POL( n__f1(x1) ) = 3x1 + 1


POL( F1(x1) ) = 2x1 + 1



The following usable rules [14] were oriented: none



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

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

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

The TRS R consists of the following rules:

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