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

(0) Obligation:

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

+(0, y) → y
+(s(x), y) → s(+(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, 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))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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(s(x), s(y)) → -1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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 2 SCCs.

(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:

+(0, y) → y
+(s(x), y) → s(+(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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.


-1(s(x), s(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)  =  x2
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)
0  =  0
-(x1, x2)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
+2 > s1

Status:
s1: [1]
+2: [1,2]
0: []


The following usable rules [FROCOS05] were oriented:

+(0, y) → y
+(s(x), y) → s(+(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, 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))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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(s(x), y) → +1(x, y)

The TRS R consists of the following rules:

+(0, y) → y
+(s(x), y) → s(+(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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(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
s(x1)  =  s(x1)
+(x1, x2)  =  +(x1, x2)
0  =  0
-(x1, x2)  =  x1

Lexicographic path order with status [LPO].
Quasi-Precedence:
+2 > s1

Status:
s1: [1]
+2: [1,2]
0: []


The following usable rules [FROCOS05] were oriented:

+(0, y) → y
+(s(x), y) → s(+(x, y))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, 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))
-(0, y) → 0
-(x, 0) → x
-(s(x), s(y)) → -(x, y)

The set Q consists of the following terms:

+(0, x0)
+(s(x0), x1)
-(0, x0)
-(x0, 0)
-(s(x0), s(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