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:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

Q is empty.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

G(x, c(y)) → G(x, g(s(c(y)), y))
F(s(x)) → F(x)
G(s(x), s(y)) → IF(f(x), s(x), s(y))
G(x, c(y)) → G(s(c(y)), y)
G(s(x), s(y)) → F(x)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(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:

G(x, c(y)) → G(x, g(s(c(y)), y))
F(s(x)) → F(x)
G(s(x), s(y)) → IF(f(x), s(x), s(y))
G(x, c(y)) → G(s(c(y)), y)
G(s(x), s(y)) → F(x)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

F(s(x)) → F(x)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


F(s(x)) → F(x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1/4 + (7/2)x_1   
POL(F(x1)) = (2)x_1   
The value of delta used in the strict ordering is 1/2.
The following usable rules [17] 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:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(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:

G(x, c(y)) → G(x, g(s(c(y)), y))
G(x, c(y)) → G(s(c(y)), y)

The TRS R consists of the following rules:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(y)), y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


G(x, c(y)) → G(x, g(s(c(y)), y))
G(x, c(y)) → G(s(c(y)), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(c(x1)) = 5/4 + (5/4)x_1   
POL(f(x1)) = 4 + (2)x_1   
POL(if(x1, x2, x3)) = 3/4 + (3/2)x_2 + (2)x_3   
POL(g(x1, x2)) = 3/4   
POL(true) = 15/4   
POL(false) = 3   
POL(s(x1)) = 0   
POL(G(x1, x2)) = (1/2)x_2   
POL(0) = 4   
POL(1) = 5/4   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(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:

f(0) → true
f(1) → false
f(s(x)) → f(x)
if(true, x, y) → x
if(false, x, y) → y
g(s(x), s(y)) → if(f(x), s(x), s(y))
g(x, c(y)) → g(x, g(s(c(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.