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

(0) Obligation:

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

p(s(x)) → x
s(p(x)) → x
+(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) 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), y) → S(+(x, y))
+1(s(x), y) → +1(x, y)
+1(p(x), y) → P(+(x, y))
+1(p(x), y) → +1(x, y)
MINUS(s(x)) → P(minus(x))
MINUS(s(x)) → MINUS(x)
MINUS(p(x)) → S(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:

p(s(x)) → x
s(p(x)) → x
+(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.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 7 less nodes.

(4) Complex Obligation (AND)

(5) 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:

p(s(x)) → x
s(p(x)) → x
+(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.
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.


MINUS(p(x)) → MINUS(x)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
MINUS(x0, x1)  =  MINUS(x1)

Tags:
MINUS has argument tags [0,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(MINUS(x1)) = 1   
POL(p(x1)) = 1 + x1   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

MINUS(s(x)) → MINUS(x)

The TRS R consists of the following rules:

p(s(x)) → x
s(p(x)) → x
+(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.
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: SCNP Order with the following components:
Level mapping:
Top level AFS:
MINUS(x0, x1)  =  MINUS(x1)

Tags:
MINUS has argument tags [1,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(MINUS(x1)) = 0   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

p(s(x)) → x
s(p(x)) → x
+(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.
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:

p(s(x)) → x
s(p(x)) → x
+(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.
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)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
+1(x0, x1, x2)  =  +1(x1)

Tags:
+1 has argument tags [2,1,2] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(+1(x1, x2)) = x2   
POL(p(x1)) = 1 + x1   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(14) Obligation:

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

+1(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

p(s(x)) → x
s(p(x)) → x
+(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.
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: SCNP Order with the following components:
Level mapping:
Top level AFS:
+1(x0, x1, x2)  =  +1(x0)

Tags:
+1 has argument tags [3,1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(+1(x1, x2)) = 1 + x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

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

p(s(x)) → x
s(p(x)) → x
+(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.
We have to consider all minimal (P,Q,R)-chains.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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:

p(s(x)) → x
s(p(x)) → x
+(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.
We have to consider all minimal (P,Q,R)-chains.

(20) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
*1(x0, x1, x2)  =  *1(x1)

Tags:
*1 has argument tags [2,1,2] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(*1(x1, x2)) = x2   
POL(p(x1)) = 1 + x1   
POL(s(x1)) = x1   

The following usable rules [FROCOS05] were oriented: none

(21) Obligation:

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

*1(s(x), y) → *1(x, y)

The TRS R consists of the following rules:

p(s(x)) → x
s(p(x)) → x
+(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.
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: SCNP Order with the following components:
Level mapping:
Top level AFS:
*1(x0, x1, x2)  =  *1(x0)

Tags:
*1 has argument tags [3,1,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Polynomial interpretation [POLO]:

POL(*1(x1, x2)) = 1 + x1   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(23) Obligation:

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

p(s(x)) → x
s(p(x)) → x
+(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.
We have to consider all minimal (P,Q,R)-chains.

(24) PisEmptyProof (EQUIVALENT transformation)

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

(25) TRUE