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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 DependencyGraphProof (⇔)
↳6 AND
  ↳7 QDP
    ↳8 UsableRulesProof (⇔)
    ↳9 QDP
    ↳10 QReductionProof (⇔)
    ↳11 QDP
    ↳12 QDPSizeChangeProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 UsableRulesProof (⇔)
    ↳16 QDP
    ↳17 QReductionProof (⇔)
    ↳18 QDP
    ↳19 QDPSizeChangeProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 UsableRulesProof (⇔)
    ↳23 QDP
    ↳24 QReductionProof (⇔)
    ↳25 QDP
    ↳26 QDPSizeChangeProof (⇔)
    ↳27 TRUE
  ↳28 QDP
    ↳29 UsableRulesProof (⇔)
    ↳30 QDP
    ↳31 QReductionProof (⇔)
    ↳32 QDP
    ↳33 QDPSizeChangeProof (⇔)
    ↳34 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

+1(s(x), s(y)) → +1(x, y)
*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)
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:

+(x, 0) → x
+(0, x) → x
+(s(x), s(y)) → s(s(+(x, y)))
*(x, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

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

(6) Complex Obligation (AND)

(7) 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

(8) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) Obligation:

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

+1(s(x), s(y)) → +1(x, y)

R is empty.
The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

(10) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

(11) Obligation:

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

+1(s(x), s(y)) → +1(x, y)

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

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

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

  • +1(s(x), s(y)) → +1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(13) TRUE

(14) 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

(15) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

SUM(cons(x, l)) → SUM(l)

R is empty.
The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

(17) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

(18) Obligation:

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

SUM(cons(x, l)) → SUM(l)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

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

  • SUM(cons(x, l)) → SUM(l)
    The graph contains the following edges 1 > 1

(20) TRUE

(21) 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) Obligation:

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

*1(s(x), s(y)) → *1(x, y)

R is empty.
The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

(25) Obligation:

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

*1(s(x), s(y)) → *1(x, y)

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

(26) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

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

  • *1(s(x), s(y)) → *1(x, y)
    The graph contains the following edges 1 > 1, 2 > 2

(27) TRUE

(28) 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, 0) → 0
*(0, x) → 0
*(s(x), s(y)) → s(+(*(x, y), +(x, y)))
sum(nil) → 0
sum(cons(x, l)) → +(x, sum(l))
prod(nil) → s(0)
prod(cons(x, l)) → *(x, prod(l))

The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

(29) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(30) Obligation:

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

PROD(cons(x, l)) → PROD(l)

R is empty.
The set Q consists of the following terms:

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

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

(31) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

+(x0, 0)
+(0, x0)
+(s(x0), s(x1))
*(x0, 0)
*(0, x0)
*(s(x0), s(x1))
sum(nil)
sum(cons(x0, x1))
prod(nil)
prod(cons(x0, x1))

(32) Obligation:

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

PROD(cons(x, l)) → PROD(l)

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

(33) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

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

  • PROD(cons(x, l)) → PROD(l)
    The graph contains the following edges 1 > 1

(34) TRUE