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 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 TRUE
    ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE
    ↳14 QDP
        ↳15 QDPSizeChangeProof (⇔)
        ↳16 TRUE
    ↳17 QDP
        ↳18 QDPSizeChangeProof (⇔)
        ↳19 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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
cons(x1, x2)  =  cons(x2)

From the DPs we obtained the following set of size-change graphs:

  • APP(cons(x, l1), l2) → APP(l1, l2) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)

From the DPs we obtained the following set of size-change graphs:

  • +1(+(x, y), z) → +1(x, +(y, z)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • +1(s(x), s(y)) → +1(x, y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 > 2

  • +1(+(x, y), z) → +1(y, z) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE

(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:

+(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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  cons(x2)

From the DPs we obtained the following set of size-change graphs:

  • SUM(app(l1, l2)) → SUM(l1) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • SUM(cons(x, l)) → SUM(l) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • SUM(app(l1, l2)) → SUM(l2) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(13) TRUE

(14) 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.

(15) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)
*(x1, x2)  =  *(x1, x2)

From the DPs we obtained the following set of size-change graphs:

  • *1(*(x, y), z) → *1(x, *(y, z)) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • *1(s(x), s(y)) → *1(x, y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 > 2

  • *1(*(x, y), z) → *1(y, z) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(16) TRUE

(17) 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.

(18) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
app(x1, x2)  =  app(x1, x2)
cons(x1, x2)  =  cons(x2)

From the DPs we obtained the following set of size-change graphs:

  • PROD(app(l1, l2)) → PROD(l1) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • PROD(cons(x, l)) → PROD(l) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • PROD(app(l1, l2)) → PROD(l2) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(19) TRUE