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 QDPOrderProof (⇔)
    ↳9 QDP
    ↳10 QDPOrderProof (⇔)
    ↳11 QDP
    ↳12 PisEmptyProof (⇔)
    ↳13 TRUE
  ↳14 QDP
    ↳15 QDPOrderProof (⇔)
    ↳16 QDP
    ↳17 QDPOrderProof (⇔)
    ↳18 QDP
    ↳19 PisEmptyProof (⇔)
    ↳20 TRUE
  ↳21 QDP
    ↳22 QDPOrderProof (⇔)
    ↳23 QDP
    ↳24 QDPOrderProof (⇔)
    ↳25 QDP
    ↳26 PisEmptyProof (⇔)
    ↳27 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

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:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(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), y) → +1(x, y)
+1(p(x), y) → +1(x, y)
MINUS(s(x)) → MINUS(x)
MINUS(p(x)) → MINUS(x)
*1(s(x), y) → +1(*(x, y), y)
*1(s(x), y) → *1(x, y)
*1(p(x), y) → +1(*(x, y), minus(y))
*1(p(x), y) → *1(x, y)
*1(p(x), y) → MINUS(y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(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 3 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

MINUS(p(x)) → MINUS(x)
MINUS(s(x)) → MINUS(x)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(s(x)) → MINUS(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1)  =  x1
p(x1)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

MINUS(p(x)) → MINUS(x)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

(10) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(p(x)) → MINUS(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1)  =  x1
p(x1)  =  p(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) Obligation:

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

+1(p(x), y) → +1(x, y)
+1(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


+1(s(x), 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)  =  x1
p(x1)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

+1(p(x), y) → +1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

(17) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


+1(p(x), 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)  =  x1
p(x1)  =  p(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(18) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

(19) PisEmptyProof (EQUIVALENT transformation)

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

(20) TRUE

(21) Obligation:

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

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

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

(22) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


*1(s(x), 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)  =  x1
p(x1)  =  x1
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

*1(p(x), y) → *1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

(24) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


*1(p(x), 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)  =  x1
p(x1)  =  p(x1)

Recursive Path Order [RPO].
Precedence:
trivial


The following usable rules [FROCOS05] were oriented: none

(25) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
+(p(x), y) → p(+(x, y))
minus(0) → 0
minus(s(x)) → p(minus(x))
minus(p(x)) → s(minus(x))
*(0, y) → 0
*(s(x), y) → +(*(x, y), y)
*(p(x), y) → +(*(x, y), minus(y))

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
+(p(x0), x1)
minus(0)
minus(s(x0))
minus(p(x0))
*(0, x0)
*(s(x0), x1)
*(p(x0), x1)

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

(26) PisEmptyProof (EQUIVALENT transformation)

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

(27) TRUE