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(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, x, y) -> x
if3(false, x, y) -> y
g2(s1(x), s1(y)) -> if3(f1(x), s1(x), s1(y))
g2(x, c1(y)) -> g2(x, g2(s1(c1(y)), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, x, y) -> x
if3(false, x, y) -> y
g2(s1(x), s1(y)) -> if3(f1(x), s1(x), s1(y))
g2(x, c1(y)) -> g2(x, g2(s1(c1(y)), y))

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:

G2(x, c1(y)) -> G2(x, g2(s1(c1(y)), y))
G2(s1(x), s1(y)) -> IF3(f1(x), s1(x), s1(y))
G2(s1(x), s1(y)) -> F1(x)
G2(x, c1(y)) -> G2(s1(c1(y)), y)
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, x, y) -> x
if3(false, x, y) -> y
g2(s1(x), s1(y)) -> if3(f1(x), s1(x), s1(y))
g2(x, c1(y)) -> g2(x, g2(s1(c1(y)), y))

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:

G2(x, c1(y)) -> G2(x, g2(s1(c1(y)), y))
G2(s1(x), s1(y)) -> IF3(f1(x), s1(x), s1(y))
G2(s1(x), s1(y)) -> F1(x)
G2(x, c1(y)) -> G2(s1(c1(y)), y)
F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, x, y) -> x
if3(false, x, y) -> y
g2(s1(x), s1(y)) -> if3(f1(x), s1(x), s1(y))
g2(x, c1(y)) -> g2(x, g2(s1(c1(y)), y))

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

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

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

F1(s1(x)) -> F1(x)

The TRS R consists of the following rules:

f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, x, y) -> x
if3(false, x, y) -> y
g2(s1(x), s1(y)) -> if3(f1(x), s1(x), s1(y))
g2(x, c1(y)) -> g2(x, g2(s1(c1(y)), y))

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(s1(x)) -> F1(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(F1(x1)) = x1   
POL(s1(x1)) = 1 + x1   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof
          ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, x, y) -> x
if3(false, x, y) -> y
g2(s1(x), s1(y)) -> if3(f1(x), s1(x), s1(y))
g2(x, c1(y)) -> g2(x, g2(s1(c1(y)), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

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

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

G2(x, c1(y)) -> G2(x, g2(s1(c1(y)), y))
G2(x, c1(y)) -> G2(s1(c1(y)), y)

The TRS R consists of the following rules:

f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, x, y) -> x
if3(false, x, y) -> y
g2(s1(x), s1(y)) -> if3(f1(x), s1(x), s1(y))
g2(x, c1(y)) -> g2(x, g2(s1(c1(y)), y))

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.


G2(x, c1(y)) -> G2(x, g2(s1(c1(y)), y))
G2(x, c1(y)) -> G2(s1(c1(y)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(1) = 0   
POL(G2(x1, x2)) = x1 + x2   
POL(c1(x1)) = 1 + x1   
POL(f1(x1)) = 1 + x1   
POL(false) = 0   
POL(g2(x1, x2)) = x1 + x2   
POL(if3(x1, x2, x3)) = x2 + x3   
POL(s1(x1)) = 0   
POL(true) = 0   

The following usable rules [14] were oriented:

g2(s1(x), s1(y)) -> if3(f1(x), s1(x), s1(y))
if3(true, x, y) -> x
g2(x, c1(y)) -> g2(x, g2(s1(c1(y)), y))
if3(false, x, y) -> y



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPOrderProof
              ↳ QDP
                ↳ PisEmptyProof

Q DP problem:
P is empty.
The TRS R consists of the following rules:

f1(0) -> true
f1(1) -> false
f1(s1(x)) -> f1(x)
if3(true, x, y) -> x
if3(false, x, y) -> y
g2(s1(x), s1(y)) -> if3(f1(x), s1(x), s1(y))
g2(x, c1(y)) -> g2(x, g2(s1(c1(y)), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.