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 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 QDPOrderProof (⇔)
    ↳14 QDP
    ↳15 PisEmptyProof (⇔)
    ↳16 TRUE
  ↳17 QDP
    ↳18 QDPOrderProof (⇔)
    ↳19 QDP
    ↳20 PisEmptyProof (⇔)
    ↳21 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(p(x)) → MINUS(x)
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)  =  p(x1)
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)
0  =  0
minus(x1)  =  minus(x1)
*(x1, x2)  =  *(x1, x2)

Recursive Path Order [RPO].
Precedence:
0 > p1
*2 > +2 > s1 > p1
*2 > minus1 > s1 > p1

The following usable rules [FROCOS05] were oriented:

+(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))

(9) 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.

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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.

(13) 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)
+1(s(x), y) → +1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
+^12 > p1
*2 > +2 > s1 > p1
*2 > minus1 > s1 > p1
*2 > minus1 > 0 > p1

The following usable rules [FROCOS05] were oriented:

+(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))

(14) 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.

(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:

*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.

(18) 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)
*1(s(x), y) → *1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Recursive Path Order [RPO].
Precedence:
*^12 > p1
*2 > +2 > s1 > p1
*2 > minus1 > s1 > p1
*2 > minus1 > 0 > p1

The following usable rules [FROCOS05] were oriented:

+(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))

(19) 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.

(20) PisEmptyProof (EQUIVALENT transformation)

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

(21) TRUE