Termination w.r.t. Q of the given QTRS could be proven:

(0) Obligation:

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

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

+¹(0(x), 0(y)) → 0¹(+(x, y))
+¹(0(x), 0(y)) → +¹(x, y)
+¹(0(x), 1(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(1(x), 1(y)) → 0¹(+(+(x, y), 1(#)))
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
+¹(1(x), 1(y)) → +¹(x, y)
+¹(+(x, y), z) → +¹(x, +(y, z))
+¹(+(x, y), z) → +¹(y, z)
*¹(0(x), y) → 0¹(*(x, y))
*¹(0(x), y) → *¹(x, y)
*¹(1(x), y) → +¹(0(*(x, y)), y)
*¹(1(x), y) → 0¹(*(x, y))
*¹(1(x), y) → *¹(x, y)
*¹(*(x, y), z) → *¹(x, *(y, z))
*¹(*(x, y), z) → *¹(y, z)
SUM(nil) → 0¹(#)
SUM(cons(x, l)) → +¹(x, sum(l))
SUM(cons(x, l)) → SUM(l)
PROD(cons(x, l)) → *¹(x, prod(l))
PROD(cons(x, l)) → PROD(l)

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 8 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

+¹(0(x), 1(y)) → +¹(x, y)
+¹(0(x), 0(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
+¹(1(x), 1(y)) → +¹(x, y)
+¹(+(x, y), z) → +¹(x, +(y, z))
+¹(+(x, y), z) → +¹(y, z)

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

+¹(0(x), 1(y)) → +¹(x, y)
+¹(1(x), 0(y)) → +¹(x, y)
+¹(1(x), 1(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
+¹(x0, x1, x2)  =  +¹(x0, x1)

Tags:
+¹ has argument tags [0,2,3] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(#) = 1   
POL(+(x₁, x₂)) = x₁ + x₂   
POL(+¹(x₁, x₂)) = x₁ + x₂   
POL(0(x₁)) = x₁   
POL(1(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented:

+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
0(#) → #

(7) Obligation:

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

+¹(0(x), 0(y)) → +¹(x, y)
+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))
+¹(+(x, y), z) → +¹(x, +(y, z))
+¹(+(x, y), z) → +¹(y, z)

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

+¹(1(x), 1(y)) → +¹(+(x, y), 1(#))

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
+¹(x0, x1, x2)  =  +¹(x0, x1, x2)

Tags:
+¹ has argument tags [0,3,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(#) = 0   
POL(+(x₁, x₂)) = x₁ + x₂   
POL(+¹(x₁, x₂)) = x₁ + x₂   
POL(0(x₁)) = x₁   
POL(1(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented:

+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
0(#) → #

(9) Obligation:

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

+¹(0(x), 0(y)) → +¹(x, y)
+¹(+(x, y), z) → +¹(x, +(y, z))
+¹(+(x, y), z) → +¹(y, z)

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

+¹(0(x), 0(y)) → +¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
+¹(x0, x1, x2)  =  +¹(x1)

Tags:
+¹ has argument tags [0,0,3] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(#) = 0   
POL(+(x₁, x₂)) = x₁ + x₂   
POL(+¹(x₁, x₂)) = 0   
POL(0(x₁)) = 1 + x₁   
POL(1(x₁)) = 0   

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

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

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

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(12) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
+¹(x0, x1, x2)  =  +¹(x1)

Tags:
+¹ has argument tags [2,0,2] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(#) = 0   
POL(+(x₁, x₂)) = 1 + x₁ + x₂   
POL(+¹(x₁, x₂)) = x₂   
POL(0(x₁)) = 0   
POL(1(x₁)) = 0   

The following usable rules [FROCOS05] were oriented:

+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
0(#) → #

(13) Obligation:

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

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(14) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(15) TRUE

(16) Obligation:

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:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

SUM(cons(x, l)) → SUM(l)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
SUM(x0, x1)  =  SUM(x1)

Tags:
SUM has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(SUM(x₁)) = 1   
POL(cons(x₁, x₂)) = 1 + x₂   

The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

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

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(19) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(20) TRUE

(21) Obligation:

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

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

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

*¹(0(x), y) → *¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
*¹(x0, x1, x2)  =  *¹(x1)

Tags:
*¹ has argument tags [1,0,2] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(#) = 1   
POL(*(x₁, x₂)) = x₁ + x₂   
POL(*¹(x₁, x₂)) = 1   
POL(+(x₁, x₂)) = 1 + x₂   
POL(0(x₁)) = 1 + x₁   
POL(1(x₁)) = x₁   

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

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

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

*¹(1(x), y) → *¹(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
*¹(x0, x1, x2)  =  *¹(x1)

Tags:
*¹ has argument tags [0,0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(#) = 1   
POL(*(x₁, x₂)) = x₁ + x₂   
POL(*¹(x₁, x₂)) = 0   
POL(+(x₁, x₂)) = 1 + x₂   
POL(0(x₁)) = 0   
POL(1(x₁)) = 1 + x₁   

The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

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

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

The TRS R consists of the following rules:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(26) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

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

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
*¹(x0, x1, x2)  =  *¹(x1)

Tags:
*¹ has argument tags [1,3,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(#) = 1   
POL(*(x₁, x₂)) = 1 + x₁ + x₂   
POL(*¹(x₁, x₂)) = 0   
POL(+(x₁, x₂)) = x₂   
POL(0(x₁)) = 1 + x₁   
POL(1(x₁)) = x₁   

The following usable rules [FROCOS05] were oriented: none

(27) Obligation:

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

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(28) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(29) TRUE

(30) Obligation:

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:

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(31) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

PROD(cons(x, l)) → PROD(l)

The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
PROD(x0, x1)  =  PROD(x1)

Tags:
PROD has argument tags [1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(PROD(x₁)) = 1   
POL(cons(x₁, x₂)) = 1 + x₂   

The following usable rules [FROCOS05] were oriented: none

(32) Obligation:

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

0(#) → #
+(x, #) → x
+(#, x) → x
+(0(x), 0(y)) → 0(+(x, y))
+(0(x), 1(y)) → 1(+(x, y))
+(1(x), 0(y)) → 1(+(x, y))
+(1(x), 1(y)) → 0(+(+(x, y), 1(#)))
+(+(x, y), z) → +(x, +(y, z))
*(#, x) → #
*(0(x), y) → 0(*(x, y))
*(1(x), y) → +(0(*(x, y)), y)
*(*(x, y), z) → *(x, *(y, z))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))

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

(33) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(34) TRUE