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:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(true, x, y) → P(y)
COND2(false, x, y) → P(x)
COND1(true, x, y) → GR(y, 0)
COND2(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND2(false, x, y) → GR(x, 0)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(true, x, y) → P(y)
COND2(false, x, y) → P(x)
COND1(true, x, y) → GR(y, 0)
COND2(true, x, y) → GR(y, 0)
GR(s(x), s(y)) → GR(x, y)
COND2(false, x, y) → GR(x, 0)
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

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 5 less nodes.

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

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

GR(s(x), s(y)) → GR(x, y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

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.


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

POL(GR(x1, x2)) = x_2   
POL(s(x1)) = 1 + x_1   
The value of delta used in the strict ordering is 1.
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:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

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:

COND1(true, x, y) → COND2(gr(y, 0), x, y)
COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

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.


COND1(true, x, y) → COND2(gr(y, 0), x, y)
The remaining pairs can at least be oriented weakly.

COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)
Used ordering: Polynomial interpretation [25,35]:

POL(gr(x1, x2)) = (1/4)x_1   
POL(true) = 1/4   
POL(COND2(x1, x2, x3)) = x_2   
POL(false) = 0   
POL(p(x1)) = (1/2)x_1   
POL(s(x1)) = 1 + (2)x_1   
POL(0) = 0   
POL(COND1(x1, x2, x3)) = (1/4)x_1 + x_2   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented:

p(s(x)) → x
p(0) → 0
gr(0, x) → false
gr(s(x), 0) → true



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

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

COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
COND2(false, x, y) → COND1(gr(x, 0), p(x), y)

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

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

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

COND2(true, x, y) → COND2(gr(y, 0), x, p(y))

The TRS R consists of the following rules:

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

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.


COND2(true, x, y) → COND2(gr(y, 0), x, p(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(gr(x1, x2)) = (1/4)x_1   
POL(COND2(x1, x2, x3)) = (1/4)x_1 + (1/2)x_3   
POL(true) = 1   
POL(false) = 0   
POL(p(x1)) = (1/4)x_1   
POL(s(x1)) = 4 + (4)x_1   
POL(0) = 0   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented:

p(s(x)) → x
p(0) → 0
gr(0, x) → false
gr(s(x), 0) → true



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

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

cond1(true, x, y) → cond2(gr(y, 0), x, y)
cond2(true, x, y) → cond2(gr(y, 0), x, p(y))
cond2(false, x, y) → cond1(gr(x, 0), p(x), y)
gr(0, x) → false
gr(s(x), 0) → true
gr(s(x), s(y)) → gr(x, y)
p(0) → 0
p(s(x)) → x

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.