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 PisEmptyProof (⇔)
    ↳14 TRUE
  ↳15 QDP
    ↳16 QDPOrderProof (⇔)
    ↳17 QDP
    ↳18 PisEmptyProof (⇔)
    ↳19 TRUE
  ↳20 QDP
    ↳21 QDPOrderProof (⇔)
    ↳22 QDP
    ↳23 PisEmptyProof (⇔)
    ↳24 TRUE
  ↳25 QDP
    ↳26 QDPOrderProof (⇔)
    ↳27 QDP
    ↳28 PisEmptyProof (⇔)
    ↳29 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)  =  APP(x1, x2)
cons(x1, x2)  =  cons(x1, x2)
+(x1, x2)  =  +(x1, x2)
0  =  0
s(x1)  =  s
*(x1, x2)  =  x2
app(x1, x2)  =  app(x1, x2)
nil  =  nil
sum(x1)  =  sum(x1)
prod(x1)  =  prod

Lexicographic path order with status [LPO].
Quasi-Precedence:
APP2 > [0, nil]
app2 > cons2 > [0, nil]
sum1 > +2 > [s, prod] > [0, nil]

Status:
APP2: [2,1]
cons2: [2,1]
+2: [1,2]
0: []
s: []
app2: [2,1]
nil: []
sum1: [1]
prod: []


The following usable rules [FROCOS05] were oriented:

+(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))

(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(s(x), s(y)) → +1(x, y)
+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)  =  x1
+(x1, x2)  =  +(x1, x2)
s(x1)  =  s(x1)
0  =  0
*(x1, x2)  =  *(x1, x2)
app(x1, x2)  =  app(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sum(x1)  =  sum(x1)
prod(x1)  =  prod(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
[app2, cons2, sum1] > *2 > +2 > [s1, 0, nil, prod1]

Status:
+2: [1,2]
s1: [1]
0: []
*2: [1,2]
app2: [1,2]
nil: []
cons2: [1,2]
sum1: [1]
prod1: [1]


The following usable rules [FROCOS05] were oriented:

+(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))

(12) 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.

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) 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.

(16) 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(cons(x, l)) → SUM(l)
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)  =  cons(x1, x2)
+(x1, x2)  =  +(x1, x2)
0  =  0
s(x1)  =  s
*(x1, x2)  =  x2
nil  =  nil
sum(x1)  =  x1
prod(x1)  =  prod

Lexicographic path order with status [LPO].
Quasi-Precedence:
app2 > cons2 > +2 > [0, nil]
[s, prod] > +2 > [0, nil]

Status:
app2: [2,1]
cons2: [2,1]
+2: [1,2]
0: []
s: []
nil: []
prod: []


The following usable rules [FROCOS05] were oriented:

+(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))

(17) 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.

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE

(20) 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.

(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, z))
*1(s(x), s(y)) → *1(x, y)
*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, x2)
*(x1, x2)  =  *(x1, x2)
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)
0  =  0
app(x1, x2)  =  app(x1, x2)
nil  =  nil
cons(x1, x2)  =  cons(x1, x2)
sum(x1)  =  sum(x1)
prod(x1)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
*^12 > *2 > +2 > s1
*^12 > *2 > 0
app2 > cons2 > *2 > +2 > s1
app2 > cons2 > *2 > 0
app2 > cons2 > sum1 > +2 > s1
app2 > cons2 > sum1 > 0
nil > s1
nil > 0

Status:
*^12: [1,2]
*2: [1,2]
s1: [1]
+2: [1,2]
0: []
app2: [2,1]
nil: []
cons2: [1,2]
sum1: [1]


The following usable rules [FROCOS05] were oriented:

+(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))

(22) 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.

(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:

+(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.

(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(cons(x, l)) → PROD(l)
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)  =  cons(x1, x2)
+(x1, x2)  =  +(x1, x2)
0  =  0
s(x1)  =  s
*(x1, x2)  =  x2
nil  =  nil
sum(x1)  =  x1
prod(x1)  =  prod

Lexicographic path order with status [LPO].
Quasi-Precedence:
app2 > cons2 > +2 > [0, nil]
[s, prod] > +2 > [0, nil]

Status:
app2: [2,1]
cons2: [2,1]
+2: [1,2]
0: []
s: []
nil: []
prod: []


The following usable rules [FROCOS05] were oriented:

+(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))

(27) 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.

(28) PisEmptyProof (EQUIVALENT transformation)

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

(29) TRUE