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:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, y))

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, 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:

+1(x, s(y)) → +1(x, y)
ODD(s(x)) → ODD(x)
+1(s(x), y) → +1(x, y)
ODD(s(x)) → NOT(odd(x))

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, 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:

+1(x, s(y)) → +1(x, y)
ODD(s(x)) → ODD(x)
+1(s(x), y) → +1(x, y)
ODD(s(x)) → NOT(odd(x))

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, 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 1 less node.

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

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

+1(x, s(y)) → +1(x, y)
+1(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, 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.


+1(s(x), y) → +1(x, y)
The remaining pairs can at least be oriented weakly.

+1(x, s(y)) → +1(x, y)
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1/4 + x_1   
POL(+1(x1, x2)) = x_1   
The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none



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

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

+1(x, s(y)) → +1(x, y)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, 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.


+1(x, s(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 1/4 + (2)x_1   
POL(+1(x1, x2)) = (1/4)x_2   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



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

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, 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:

ODD(s(x)) → ODD(x)

The TRS R consists of the following rules:

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, 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.


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

POL(ODD(x1)) = (1/4)x_1   
POL(s(x1)) = 1/4 + (2)x_1   
The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none



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

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

not(true) → false
not(false) → true
odd(0) → false
odd(s(x)) → not(odd(x))
+(x, 0) → x
+(x, s(y)) → s(+(x, y))
+(s(x), y) → s(+(x, 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.