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

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

(0) Obligation:

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))
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) → s(0)
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(s(x), s(y)) → +1(x, y)
+1(+(x, y), z) → +1(x, +(y, z))
+1(+(x, y), z) → +1(y, z)
*1(s(x), s(y)) → +1(*(x, y), +(x, y))
*1(s(x), s(y)) → *1(x, y)
*1(s(x), s(y)) → +1(x, y)
*1(*(x, y), z) → *1(x, *(y, z))
*1(*(x, y), z) → *1(y, z)
APP(cons(x, l1), l2) → APP(l1, l2)
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:

+(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))
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) → s(0)
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 6 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:

+(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))
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) → s(0)
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)

Lexicographic Path Order [LPO].
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:

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

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))
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) → s(0)
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) 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: Combined order from the following AFS and order.
+1(x1, x2)  =  +1(x1)
+(x1, x2)  =  +(x1, x2)
s(x1)  =  x1
0  =  0

Lexicographic Path Order [LPO].
Precedence:
+^11 > +2


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

+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))
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) → s(0)
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.

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


+1(s(x), s(y)) → +1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
+1(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

Q DP problem:
P is empty.
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))
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) → s(0)
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.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE

(17) 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:

+(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))
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) → s(0)
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.

(18) 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)  =  SUM(x1)
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  x2

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(19) 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:

+(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))
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) → s(0)
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.

(20) 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(x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

Q DP problem:
P is empty.
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))
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) → s(0)
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.

(22) PisEmptyProof (EQUIVALENT transformation)

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

(23) TRUE

(24) Obligation:

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

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

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))
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) → s(0)
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.

(25) 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: Combined order from the following AFS and order.
*1(x1, x2)  =  *1(x1)
*(x1, x2)  =  *(x1, x2)
s(x1)  =  x1
+(x1, x2)  =  +(x2)
0  =  0

Lexicographic Path Order [LPO].
Precedence:
*^11 > *2
+1 > *2
0 > *2


The following usable rules [FROCOS05] were oriented: none

(26) Obligation:

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

*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))
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) → s(0)
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.

(27) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


*1(s(x), s(y)) → *1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
*1(x1, x2)  =  x2
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(28) Obligation:

Q DP problem:
P is empty.
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))
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) → s(0)
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.

(29) PisEmptyProof (EQUIVALENT transformation)

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

(30) TRUE

(31) 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:

+(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))
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) → s(0)
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.

(32) 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)  =  PROD(x1)
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  x2

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(33) 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:

+(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))
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) → s(0)
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.

(34) 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(x2)

Lexicographic Path Order [LPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(35) Obligation:

Q DP problem:
P is empty.
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))
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) → s(0)
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.

(36) PisEmptyProof (EQUIVALENT transformation)

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

(37) TRUE