Termination w.r.t. Q of the given QTRS could not be shown:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
  ↳11 QDP
    ↳12 QDPOrderProof (⇔)
    ↳13 QDP
    ↳14 QDPOrderProof (⇔)
    ↳15 QDP
    ↳16 PisEmptyProof (⇔)
    ↳17 TRUE
  ↳18 QDP
    ↳19 QDPOrderProof (⇔)
    ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 TRUE
  ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 QDPOrderProof (⇔)
    ↳29 QDP
    ↳30 PisEmptyProof (⇔)
    ↳31 TRUE

(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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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)
*1(x, +(y, z)) → +1(*(x, y), *(x, z))
*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, z)
APP(cons(x, l1), l2) → APP(l1, l2)
SUM(nil) → 01(#)
SUM(cons(x, l)) → +1(x, sum(l))
SUM(cons(x, l)) → SUM(l)
SUM(app(l1, l2)) → +1(sum(l1), sum(l2))
SUM(app(l1, l2)) → SUM(l1)
SUM(app(l1, l2)) → SUM(l2)
PROD(cons(x, l)) → *1(x, prod(l))
PROD(cons(x, l)) → PROD(l)
PROD(app(l1, l2)) → *1(prod(l1), prod(l2))
PROD(app(l1, l2)) → PROD(l1)
PROD(app(l1, l2)) → PROD(l2)

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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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 5 SCCs with 11 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

APP(cons(x, l1), l2) → APP(l1, l2)

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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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.


APP(cons(x, l1), l2) → APP(l1, l2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
APP(x1, x2)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(7) 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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

(11) Obligation:

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

SUM(app(l1, l2)) → SUM(l1)
SUM(cons(x, l)) → SUM(l)
SUM(app(l1, l2)) → SUM(l2)

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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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.


SUM(app(l1, l2)) → SUM(l1)
SUM(app(l1, l2)) → SUM(l2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
SUM(x1)  =  x1
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(13) 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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

(14) 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: Combined order from the following AFS and order.
SUM(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(15) 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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

(16) PisEmptyProof (EQUIVALENT transformation)

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

(17) TRUE

(18) 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)
*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, 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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

(19) 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)
*1(0(x), y) → *1(x, y)
*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: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1)
1(x1)  =  1(x1)
0(x1)  =  0(x1)
*(x1, x2)  =  *(x1, x2)
+(x1, x2)  =  x2
#  =  #

Recursive Path Order [RPO].
Precedence:
11 > 01 > [*^11, *2, #]


The following usable rules [FROCOS05] were oriented: none

(20) Obligation:

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)

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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

(21) 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)
*1(x, +(y, z)) → *1(x, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(22) 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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

(23) PisEmptyProof (EQUIVALENT transformation)

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

(24) TRUE

(25) Obligation:

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

PROD(app(l1, l2)) → PROD(l1)
PROD(cons(x, l)) → PROD(l)
PROD(app(l1, l2)) → PROD(l2)

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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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.


PROD(app(l1, l2)) → PROD(l1)
PROD(app(l1, l2)) → PROD(l2)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
PROD(x1)  =  x1
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  x2

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(27) 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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

(28) 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: Combined order from the following AFS and order.
PROD(x1)  =  x1
cons(x1, x2)  =  cons(x1, x2)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(29) 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))
*(x, +(y, z)) → +(*(x, y), *(x, z))
app(nil, l) → l
app(cons(x, l1), l2) → cons(x, app(l1, l2))
sum(nil) → 0(#)
sum(cons(x, l)) → +(x, sum(l))
sum(app(l1, l2)) → +(sum(l1), sum(l2))
prod(nil) → 1(#)
prod(cons(x, l)) → *(x, prod(l))
prod(app(l1, l2)) → *(prod(l1), prod(l2))

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

(30) PisEmptyProof (EQUIVALENT transformation)

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

(31) TRUE