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:

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x

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:

+1(x, +(y, z)) → +1(x, y)
I(+(x, y)) → I(x)
+1(x, +(y, z)) → +1(+(x, y), z)
I(+(x, y)) → I(y)
I(+(x, y)) → +1(i(x), i(y))

The TRS R consists of the following rules:

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x

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

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof

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

+1(x, +(y, z)) → +1(x, y)
I(+(x, y)) → I(x)
+1(x, +(y, z)) → +1(+(x, y), z)
I(+(x, y)) → I(y)
I(+(x, y)) → +1(i(x), i(y))

The TRS R consists of the following rules:

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof

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

I(+(x, y)) → I(x)
+1(x, +(y, z)) → +1(x, y)
I(+(x, y)) → I(y)
+1(x, +(y, z)) → +1(+(x, y), z)
I(+(x, y)) → +1(i(x), i(y))

The TRS R consists of the following rules:

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x

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 1 less node.

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

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

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

The TRS R consists of the following rules:

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x

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.


+1(x, +(y, z)) → +1(x, y)
+1(x, +(y, z)) → +1(+(x, y), z)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1, x2)
+(x1, x2)  =  +(x1, x2)
i(x1)  =  i
0  =  0

Recursive path order with status [2].
Quasi-Precedence:
+^12 > [+2, i, 0]

Status:
+2: [2,1]
0: multiset
i: multiset
+^12: [2,1]


The following usable rules [14] were oriented:

+(x, i(x)) → 0
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(0, y) → y
+(i(x), x) → 0
+(+(x, y), i(y)) → x
+(x, 0) → x



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

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

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → 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
      ↳ EdgeDeletionProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP
                ↳ QDPOrderProof

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

I(+(x, y)) → I(x)
I(+(x, y)) → I(y)

The TRS R consists of the following rules:

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → x

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.


I(+(x, y)) → I(x)
I(+(x, y)) → I(y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
I(x1)  =  x1
+(x1, x2)  =  +(x1, x2)

Recursive path order with status [2].
Quasi-Precedence:
trivial

Status:
+2: multiset


The following usable rules [14] were oriented: none



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

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

i(0) → 0
+(0, y) → y
+(x, 0) → x
i(i(x)) → x
+(i(x), x) → 0
+(x, i(x)) → 0
i(+(x, y)) → +(i(x), i(y))
+(x, +(y, z)) → +(+(x, y), z)
+(+(x, i(y)), y) → x
+(+(x, y), i(y)) → 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.