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:

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

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

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

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

The TRS R consists of the following rules:

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

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

The TRS R consists of the following rules:

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

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
            ↳ UsableRulesProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof
          ↳ QDP

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: 



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We can use the usable rules and reduction pair processor [15] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its argument. Then, we can delete all non-usable rules [17] from R.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
              ↳ QDP
                ↳ QDPSizeChangeProof

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

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

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
By using the subterm criterion [20] together with the size-change analysis [32] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs: