(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
Recursive Path Order [RPO].
Precedence:
+2 > s1
0 > s1
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) =
+1(
x1,
x2)
s(
x1) =
s(
x1)
+(
x1,
x2) =
+(
x1,
x2)
0 =
0
-(
x1,
x2) =
-(
x1)
Recursive Path Order [RPO].
Precedence:
+2 > s1
-1 > 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