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
  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

f1(0)
f1(1)
f1(s1(x0))
if3(true, x0, x1)
if3(false, x0, x1)
g2(s1(x0), s1(x1))
g2(x0, c1(x1))


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))

The set Q consists of the following terms:

f1(0)
f1(1)
f1(s1(x0))
if3(true, x0, x1)
if3(false, x0, x1)
g2(s1(x0), s1(x1))
g2(x0, c1(x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

f1(0)
f1(1)
f1(s1(x0))
if3(true, x0, x1)
if3(false, x0, x1)
g2(s1(x0), s1(x1))
g2(x0, c1(x1))

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
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

f1(0)
f1(1)
f1(s1(x0))
if3(true, x0, x1)
if3(false, x0, x1)
g2(s1(x0), s1(x1))
g2(x0, c1(x1))

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


The following pairs can be strictly oriented and are deleted.


F1(s1(x)) -> F1(x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
F1(x1)  =  F1(x1)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > F1


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

f1(0)
f1(1)
f1(s1(x0))
if3(true, x0, x1)
if3(false, x0, x1)
g2(s1(x0), s1(x1))
g2(x0, c1(x1))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

f1(0)
f1(1)
f1(s1(x0))
if3(true, x0, x1)
if3(false, x0, x1)
g2(s1(x0), s1(x1))
g2(x0, c1(x1))

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


The following pairs can be strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
G2(x1, x2)  =  G1(x2)
c1(x1)  =  c1(x1)
g2(x1, x2)  =  g1(x2)
s1(x1)  =  s
if3(x1, x2, x3)  =  if2(x2, x3)
f1(x1)  =  f
0  =  0
true  =  true
1  =  1
false  =  false

Lexicographic Path Order [19].
Precedence:
c1 > G1 > [g1, s] > if2
c1 > G1 > [g1, s] > [f, true, 1, false]


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ 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))

The set Q consists of the following terms:

f1(0)
f1(1)
f1(s1(x0))
if3(true, x0, x1)
if3(false, x0, x1)
g2(s1(x0), s1(x1))
g2(x0, c1(x1))

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