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:

a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(c) -> c
mark1(true) -> true
mark1(false) -> false
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(c) -> c
mark1(true) -> true
mark1(false) -> false
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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:

MARK1(if3(X1, X2, X3)) -> MARK1(X1)
A__F1(X) -> A__IF3(mark1(X), c, f1(true))
MARK1(f1(X)) -> A__F1(mark1(X))
A__IF3(false, X, Y) -> MARK1(Y)
A__F1(X) -> MARK1(X)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X2)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), mark1(X2), X3)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(c) -> c
mark1(true) -> true
mark1(false) -> false
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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:

MARK1(if3(X1, X2, X3)) -> MARK1(X1)
A__F1(X) -> A__IF3(mark1(X), c, f1(true))
MARK1(f1(X)) -> A__F1(mark1(X))
A__IF3(false, X, Y) -> MARK1(Y)
A__F1(X) -> MARK1(X)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X2)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), mark1(X2), X3)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(c) -> c
mark1(true) -> true
mark1(false) -> false
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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.


A__IF3(false, X, Y) -> MARK1(Y)
The remaining pairs can at least be oriented weakly.

MARK1(if3(X1, X2, X3)) -> MARK1(X1)
A__F1(X) -> A__IF3(mark1(X), c, f1(true))
MARK1(f1(X)) -> A__F1(mark1(X))
A__F1(X) -> MARK1(X)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X2)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), mark1(X2), X3)
MARK1(f1(X)) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(A__F1(x1)) = 2 + 3·x1   
POL(A__IF3(x1, x2, x3)) = 2 + 2·x1 + 3·x1·x2 + 2·x2 + 2·x2·x3 + 2·x3   
POL(MARK1(x1)) = 2 + 2·x1   
POL(a__f1(x1)) = 2·x1   
POL(a__if3(x1, x2, x3)) = x1 + 3·x1·x2 + 2·x1·x3 + 3·x2 + x2·x3 + 3·x3   
POL(c) = 0   
POL(f1(x1)) = 2·x1   
POL(false) = 1   
POL(if3(x1, x2, x3)) = x1 + 3·x1·x2 + 2·x1·x3 + 3·x2 + x2·x3 + 3·x3   
POL(mark1(x1)) = x1   
POL(true) = 0   

The following usable rules [14] were oriented:

mark1(false) -> false
mark1(true) -> true
a__if3(true, X, Y) -> mark1(X)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(f1(X)) -> a__f1(mark1(X))
a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(false, X, Y) -> mark1(Y)
mark1(c) -> c
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ QDPOrderProof
QDP
          ↳ QDPOrderProof

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

MARK1(if3(X1, X2, X3)) -> MARK1(X1)
A__F1(X) -> A__IF3(mark1(X), c, f1(true))
MARK1(f1(X)) -> A__F1(mark1(X))
A__F1(X) -> MARK1(X)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X2)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), mark1(X2), X3)
MARK1(f1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(c) -> c
mark1(true) -> true
mark1(false) -> false
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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.


MARK1(f1(X)) -> A__F1(mark1(X))
MARK1(f1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

MARK1(if3(X1, X2, X3)) -> MARK1(X1)
A__F1(X) -> A__IF3(mark1(X), c, f1(true))
A__F1(X) -> MARK1(X)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X2)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), mark1(X2), X3)
Used ordering: Polynomial interpretation [21]:

POL(A__F1(x1)) = 1 + 3·x1   
POL(A__IF3(x1, x2, x3)) = 1 + x1 + 3·x2   
POL(MARK1(x1)) = 1 + 3·x1   
POL(a__f1(x1)) = 2 + 3·x1   
POL(a__if3(x1, x2, x3)) = x1 + x1·x3 + 3·x2   
POL(c) = 0   
POL(f1(x1)) = 2 + 3·x1   
POL(false) = 2   
POL(if3(x1, x2, x3)) = x1 + x1·x3 + 3·x2   
POL(mark1(x1)) = x1   
POL(true) = 0   

The following usable rules [14] were oriented:

mark1(false) -> false
mark1(true) -> true
a__if3(true, X, Y) -> mark1(X)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(f1(X)) -> a__f1(mark1(X))
a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(false, X, Y) -> mark1(Y)
mark1(c) -> c
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)



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

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

MARK1(if3(X1, X2, X3)) -> MARK1(X1)
A__F1(X) -> A__IF3(mark1(X), c, f1(true))
A__F1(X) -> MARK1(X)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X2)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), mark1(X2), X3)

The TRS R consists of the following rules:

a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(c) -> c
mark1(true) -> true
mark1(false) -> false
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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 2 less nodes.

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

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

MARK1(if3(X1, X2, X3)) -> MARK1(X1)
A__IF3(true, X, Y) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> MARK1(X2)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), mark1(X2), X3)

The TRS R consists of the following rules:

a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(c) -> c
mark1(true) -> true
mark1(false) -> false
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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.


MARK1(if3(X1, X2, X3)) -> MARK1(X1)
MARK1(if3(X1, X2, X3)) -> MARK1(X2)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), mark1(X2), X3)
The remaining pairs can at least be oriented weakly.

A__IF3(true, X, Y) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(A__IF3(x1, x2, x3)) = 2·x2   
POL(MARK1(x1)) = x1   
POL(a__f1(x1)) = 1 + 2·x1   
POL(a__if3(x1, x2, x3)) = 1 + x1 + x1·x2 + x1·x3 + 2·x2   
POL(c) = 0   
POL(f1(x1)) = 1 + 2·x1   
POL(false) = 2   
POL(if3(x1, x2, x3)) = 1 + x1 + x1·x2 + x1·x3 + 2·x2   
POL(mark1(x1)) = x1   
POL(true) = 0   

The following usable rules [14] were oriented:

a__if3(true, X, Y) -> mark1(X)
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(f1(X)) -> a__f1(mark1(X))
a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(false, X, Y) -> mark1(Y)
mark1(c) -> c
mark1(false) -> false
a__f1(X) -> f1(X)
mark1(true) -> true
a__if3(X1, X2, X3) -> if3(X1, X2, X3)



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

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

A__IF3(true, X, Y) -> MARK1(X)

The TRS R consists of the following rules:

a__f1(X) -> a__if3(mark1(X), c, f1(true))
a__if3(true, X, Y) -> mark1(X)
a__if3(false, X, Y) -> mark1(Y)
mark1(f1(X)) -> a__f1(mark1(X))
mark1(if3(X1, X2, X3)) -> a__if3(mark1(X1), mark1(X2), X3)
mark1(c) -> c
mark1(true) -> true
mark1(false) -> false
a__f1(X) -> f1(X)
a__if3(X1, X2, X3) -> if3(X1, X2, X3)

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 1 less node.