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
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

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(s(x), s(y)) → +1(*(x, y), +(x, y))
PROD(cons(x, l)) → PROD(l)
SUM(cons(x, l)) → +1(x, sum(l))
*1(*(x, y), z) → *1(y, z)
+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
SUM(cons(x, l)) → SUM(l)
*1(*(x, y), z) → *1(x, *(y, z))
PROD(cons(x, l)) → *1(x, prod(l))
*1(s(x), s(y)) → +1(x, y)
*1(s(x), s(y)) → *1(x, y)
+1(s(x), s(y)) → +1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

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(s(x), s(y)) → +1(*(x, y), +(x, y))
PROD(cons(x, l)) → PROD(l)
SUM(cons(x, l)) → +1(x, sum(l))
*1(*(x, y), z) → *1(y, z)
+1(+(x, y), z) → +1(y, z)
+1(+(x, y), z) → +1(x, +(y, z))
SUM(cons(x, l)) → SUM(l)
*1(*(x, y), z) → *1(x, *(y, z))
PROD(cons(x, l)) → *1(x, prod(l))
*1(s(x), s(y)) → +1(x, y)
*1(s(x), s(y)) → *1(x, y)
+1(s(x), s(y)) → +1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 4 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
            ↳ UsableRulesProof
          ↳ QDP
          ↳ QDP
          ↳ 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))
+1(s(x), s(y)) → +1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

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

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(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
          ↳ QDP
          ↳ QDP

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

SUM(cons(x, l)) → SUM(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

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
          ↳ QDP
          ↳ QDP

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

SUM(cons(x, l)) → SUM(l)

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:



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
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))
*1(s(x), s(y)) → *1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

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
          ↳ QDP
QDP
            ↳ UsableRulesProof

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

PROD(cons(x, l)) → PROD(l)

The TRS R consists of the following rules:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
+(+(x, y), z) → +(x, +(y, z))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

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
          ↳ QDP
          ↳ QDP
            ↳ UsableRulesProof
QDP
                ↳ QDPSizeChangeProof

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

PROD(cons(x, l)) → PROD(l)

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: